Limits, Continuity, and Differentiation

This chapter establishes the foundational concepts of limits, continuity, and differentiability for functions of a single real variable. Mastery of these topics is crucial for the CUET PG examination, as they form the bedrock for advanced calculus and are frequently tested in various problem-solving contexts.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Limit and Continuity of Functions |
| 2 | Differentiation and its Theorems |

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We begin with Limit and Continuity of Functions.

Part 1: Limit and Continuity of Functions

We explore the foundational concepts of limits and continuity for real-valued functions of a single variable, which are critical for understanding calculus and real analysis. These principles enable us to analyze function behavior near specific points and over intervals, forming the basis for differentiation and integration.

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Core Concepts

1. Limit of a Function

We define the limit of a function $f(x)$ as $x$ approaches a point $c$ as the value $L$ that $f(x)$ gets arbitrarily close to. This definition does not require $f(c)$ to be defined or equal to $L$.

Quick Example:
Consider demonstrating $\lim_{x \to 2} (3x - 1) = 5$ using the $\epsilon-\delta$ definition.

Step 1: Start with $|f(x) - L| < \epsilon$.

Step 2: Simplify the expression.

Step 3: Isolate $|x - 2|$.

Step 4: Choose $\delta$. We choose $\delta = \frac{\epsilon}{3}$. Thus, for any $\epsilon > 0$, there exists a $\delta = \frac{\epsilon}{3}$ such that if $0 < |x - 2| < \delta$, then $|(3x - 1) - 5| < \epsilon$.

Answer: $\boxed{5}$

:::question type="MCQ" question="Given $\lim_{x \to 3} (2x+1) = 7$, which value of $\delta$ corresponds to $\epsilon = 0.01$ in the formal definition of a limit?" options=["0.005","0.01","0.02","0.0025"] answer="0.005" hint="Follow the $\epsilon-\delta$ procedure: $|(2x+1)-7| < \epsilon \implies |2x-6| < \epsilon \implies 2|x-3| < \epsilon \implies |x-3| < \epsilon/2$. Thus $\delta = \epsilon/2$." solution="Step 1: We begin with the inequality $|f(x) - L| < \epsilon$.

Step 2: Simplify the expression.

Step 3: Isolate $|x - 3|$.

Step 4: From this, we identify $\delta = \frac{\epsilon}{2}$.
Given $\epsilon = 0.01$, we calculate $\delta$.

Answer: $\boxed{0.005}$"
:::

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2. Properties of Limits

We utilize several properties to evaluate limits of combined functions, provided the individual limits exist.

Quick Example:
Evaluate $\lim_{x \to 2} (x^2 + 3x - 5)$.

Step 1: Apply the sum/difference and constant multiple rules.

Step 2: Evaluate individual limits.

Step 3: Calculate the result.

Answer: $\boxed{5}$

:::question type="MCQ" question="Given $\lim_{x \to 1} f(x) = 4$ and $\lim_{x \to 1} g(x) = -2$, determine $\lim_{x \to 1} [3f(x) - g(x)^2]$." options=["16","8","10","14"] answer="8" hint="Apply the constant multiple, difference, and power rules for limits." solution="Step 1: Apply the limit properties to the expression.

Step 2: Substitute the given limit values.

Step 3: Calculate the final value.

Answer: $\boxed{8}$"
:::

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3. One-Sided Limits

We consider the behavior of a function as $x$ approaches a point from either the left or the right. For a general limit to exist, both one-sided limits must exist and be equal.

Quick Example:
Consider the function

Evaluate $\lim_{x \to 0} f(x)$.

Step 1: Evaluate the left-hand limit.
>

>

Step 2: Evaluate the right-hand limit.
>

>

Step 3: Compare the one-sided limits. Since $\lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x)$, the limit $\lim_{x \to 0} f(x)$ does not exist.

Answer: The limit does not exist.

:::question type="MCQ" question="For the function

find $\lim_{x \to 2} f(x)$." options=["3","5","DoesNotExist","2"] answer="3" hint="Evaluate the left-hand limit and the right-hand limit separately at $x=2$. If they are equal, the limit exists and is that value." solution="Step 1: Evaluate the left-hand limit at $x=2$.
For $x < 2$, $f(x) = 2x-1$.
>
>

Step 2: Evaluate the right-hand limit at $x=2$.
For $x > 2$, $f(x) = x^2-1$.
>

>

Step 3: Compare the one-sided limits.
Since $\lim_{x \to 2^-} f(x) = 3$ and $\lim_{x \to 2^+} f(x) = 3$, the limit exists and is equal to $3$. Note that the value $f(2)=3$ is equal to the limit, but this is a condition for continuity, not for the limit's existence.

The correct option is 3."
:::

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4. Continuity at a Point

A function is continuous at a point if its limit at that point exists and equals the function's value at that point. This implies the graph of the function has no breaks, jumps, or holes at that specific point.

Quick Example:
Determine if

is continuous at $x=1$.

Step 1: Check if $f(1)$ is defined.
For $x \le 1$, $f(x) = x^2+1$.
>

$f(1)$ is defined.

Step 2: Check if $\lim_{x \to 1} f(x)$ exists.
Left-hand limit:
>

Right-hand limit:
>
Since the left-hand limit equals the right-hand limit, $\lim_{x \to 1} f(x) = 2$. The limit exists.

Step 3: Compare $\lim_{x \to 1} f(x)$ and $f(1)$.
>

>
Since $\lim_{x \to 1} f(x) = f(1)$, the function is continuous at $x=1$.

Answer: Continuous at $x=1$.

:::question type="MCQ" question="For what value of $k$ is the function

continuous at $x=1$?" options=["1","2","3","4"] answer="4" hint="For continuity at $x=1$, we must have $f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$." solution="Step 1: For $f(x)$ to be continuous at $x=1$, the function value $f(1)$, the left-hand limit $\lim_{x \to 1^-} f(x)$, and the right-hand limit $\lim_{x \to 1^+} f(x)$ must all be equal.

Step 2: Evaluate $f(1)$.
Using the first case ($x \ge 1$):
>

Step 3: Evaluate the left-hand limit.
Using the second case ($x < 1$):
>

Step 4: Evaluate the right-hand limit.
Using the first case ($x \ge 1$):
>

>

Step 5: Set the values equal for continuity.
For continuity, $k = 4$.
Thus, $f(1) = 4$, $\lim_{x \to 1^-} f(x) = 4$, and $\lim_{x \to 1^+} f(x) = 4$.

The correct option is 4."
:::

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5. Continuity on an Interval

We extend the concept of continuity from a single point to an entire interval. This implies the function's graph can be drawn without lifting the pen over that interval.

Quick Example:
Determine the interval(s) on which

is continuous.

Step 1: Identify points where the function might be undefined or have issues.
Rational functions are continuous everywhere except where the denominator is zero.

Step 2: Set the denominator to zero and solve for $x$.
>

>
The function is undefined at $x=2$.

Step 3: Conclude the intervals of continuity.
The function is continuous on $(-\infty, 2)$ and $(2, \infty)$.

Answer:

:::question type="MCQ" question="Let

On which interval is $f(x)$ continuous?" options=["No point","All points","Exactly one point","Exactly two points"] answer="Exactly one point" hint="For continuity, $f(x)$ must equal its limit. For a function like this, continuity can only occur where $x = 3-x$." solution="Step 1: For $f(x)$ to be continuous at a point $c$, we must have $\lim_{x \to c} f(x) = f(c)$.
Consider a point $c$. If $c$ is rational, $f(c) = c$. If $c$ is irrational, $f(c) = 3-c$.

Step 2: For continuity, $f(x)$ must approach $f(c)$ as $x \to c$.
If $x$ is close to $c$, $f(x)$ will take values close to $c$ (if $x$ is rational) and values close to $3-c$ (if $x$ is irrational). For the limit to exist and be equal to $f(c)$, we must have $c = 3-c$.

Step 3: Solve for $c$.
>

>
>

Step 4: Verify continuity at $c = 3/2$.
At $x = 3/2$:
$f(3/2) = 3/2$ (since $3/2$ is rational).
For any sequence of rational numbers $x_n \to 3/2$, $\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = 3/2$.
For any sequence of irrational numbers $y_n \to 3/2$, $\lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} (3-y_n) = 3 - 3/2 = 3/2$.
Since both limits are $3/2$, $\lim_{x \to 3/2} f(x) = 3/2$.
As $\lim_{x \to 3/2} f(x) = f(3/2)$, the function is continuous at $x=3/2$.

For any other point $c \ne 3/2$, the values $c$ and $3-c$ are distinct. As $x \to c$, $f(x)$ will oscillate between values near $c$ and values near $3-c$, preventing the limit from existing. Thus, $f(x)$ is continuous at exactly one point.

The correct option is Exactly one point."
:::

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6. Types of Discontinuities

Discontinuities categorize how a function fails to be continuous at a point. Understanding these types helps in analyzing function behavior.

Quick Example:
Classify the discontinuity of

at $x=2$.

Step 1: Check $f(2)$.

, which is undefined.

Step 2: Evaluate the limit as $x \to 2$.
>

>
The limit exists and is $4$.

Step 3: Classify.
Since the limit exists but $f(2)$ is undefined, this is a removable discontinuity.

Answer: Removable discontinuity.

:::question type="MCQ" question="Which of the following functions is discontinuous at every point of $\mathbb{R}$?" options=["

","","",""] answer="" hint="A function is discontinuous at every point if, for any point $c$, the limit $\lim_{x \to c} f(x)$ does not exist, or if it exists, it is not equal to $f(c)$, and this must hold for all $c \in \mathbb{R}$." solution="We analyze each option:

Option 1:

(Dirichlet function variant)
For any point $c \in \mathbb{R}$, any open interval around $c$ contains both rational and irrational numbers. Thus, $f(x)$ will take values $1$ and $-1$ arbitrarily close to $c$. The limit $\lim_{x \to c} f(x)$ does not exist for any $c$. Therefore, this function is discontinuous at every point.

Option 2:

This function is continuous only at $x=0$. If $c=0$, $f(0)=0$. As $x \to 0$, $x \to 0$ (for rational $x$) and $0 \to 0$ (for irrational $x$). So $\lim_{x \to 0} f(x) = 0 = f(0)$. At any other point $c \ne 0$, the limit does not exist because $f(x)$ oscillates between values near $c$ and $0$.

Option 3:

This function is continuous only at $x=0$. For continuity at $c$, we need $c = 2c$, which implies $c=0$. At $x=0$, $f(0)=0$, and $\lim_{x \to 0} f(x) = 0$.

Option 4:

This function is continuous only at $x=0$. For continuity at $c$, we need $c = -c$, which implies $2c=0$, so $c=0$. At $x=0$, $f(0)=0$, and $\lim_{x \to 0} f(x) = 0$.

Therefore, only the first function is discontinuous at every point of $\mathbb{R}$.

The correct option is

"
:::

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7. Properties of Continuous Functions

Continuous functions exhibit useful properties when combined, preserving continuity under various operations.

Quick Example:
Given $f(x) = x^2$ and $g(x) = \sin x$. Determine if $(f \circ g)(x)$ is continuous on $\mathbb{R}$.

Step 1: Check continuity of the inner function $g(x)$.
The sine function $g(x) = \sin x$ is continuous on $\mathbb{R}$.

Step 2: Check continuity of the outer function $f(y)$.
The function $f(y) = y^2$ (where $y=g(x)$) is a polynomial, which is continuous on $\mathbb{R}$.

Step 3: Apply the composition rule.
Since $g(x)$ is continuous on $\mathbb{R}$ and $f(y)$ is continuous on $\mathbb{R}$ (and thus at every value $g(x)$ takes), their composition $(f \circ g)(x) = f(g(x)) = (\sin x)^2$ is continuous on $\mathbb{R}$.

Answer: $(f \circ g)(x)$ is continuous on $\mathbb{R}$.

:::question type="MCQ" question="Let $f(x) = |x|$ and $g(x) = x^2 - 4$. Which of the following statements is true about the continuity of $h(x) = (f \circ g)(x)$?" options=["$h(x)$ is continuous on $\mathbb{R}$","$h(x)$ is continuous only on $(-\infty, -2) \cup (2, \infty)$","$h(x)$ is continuous only on $[-2, 2]$","$h(x)$ is discontinuous at $x=\pm 2$"] answer="$h(x)$ is continuous on $\mathbb{R}$" hint="Consider the continuity of $f(x)$ and $g(x)$ individually, then apply the composition rule. The absolute value function is continuous everywhere." solution="Step 1: Analyze the continuity of the inner function $g(x)$.
$g(x) = x^2 - 4$ is a polynomial function. Polynomials are continuous on all of $\mathbb{R}$.

Step 2: Analyze the continuity of the outer function $f(y)$.
$f(y) = |y|$ is the absolute value function. The absolute value function is continuous on all of $\mathbb{R}$.

Step 3: Apply the composition rule.
Since $g(x)$ is continuous on $\mathbb{R}$ and $f(y)$ is continuous on $\mathbb{R}$ (and therefore continuous at $g(x)$ for all $x$), their composition $h(x) = (f \circ g)(x) = |x^2 - 4|$ is continuous on all of $\mathbb{R}$. Even though $g(x)$ changes sign at $x=\pm 2$, $f(y)=|y|$ is continuous at $y=0$, which is $g(\pm 2)$.

The correct option is $h(x)$ is continuous on $\mathbb{R}$."
:::

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8. Intermediate Value Theorem (IVT)

The IVT is a fundamental theorem for continuous functions on closed intervals, guaranteeing that the function takes on every value between its values at the endpoints.

Quick Example:
Show that $f(x) = x^3 - 4x + 1$ has a root in the interval $[0, 1]$.

Step 1: Check if $f(x)$ is continuous on $[0, 1]$.
$f(x)$ is a polynomial, so it is continuous everywhere, including on $[0, 1]$.

Step 2: Evaluate $f(x)$ at the endpoints.

Step 3: Apply the IVT.
Since $f(0) = 1 > 0$ and $f(1) = -2 < 0$, and $f$ is continuous on $[0, 1]$, by the IVT, there must exist a $c \in (0, 1)$ such that $f(c) = 0$.

Answer: A root exists in $[0, 1]$.

:::question type="MCQ" question="Let $f: [0, 1] \to \mathbb{R}$ be a continuous function such that $f(0)=5$ and $f(1)=10$. Which of the following statements is guaranteed by the Intermediate Value Theorem?" options=["There exists $c \in (0, 1)$ such that $f(c)=12$","There exists $c \in (0, 1)$ such that $f(c)=7$","For all $c \in (0, 1)$, $5 < f(c) < 10$","There exists $c \in (0, 1)$ such that $f(c)=0$"] answer="There exists $c \in (0, 1)$ such that $f(c)=7$" hint="The IVT guarantees that $f(x)$ takes on every value between $f(a)$ and $f(b)$." solution="Step 1: Recall the Intermediate Value Theorem.
If $f$ is continuous on $[a, b]$, and $k$ is any number between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = k$.

Step 2: Apply the theorem to the given function.
Here $a=0$, $b=1$, $f(a)=f(0)=5$, and $f(b)=f(1)=10$.
The theorem states that for any $k$ such that $5 < k < 10$, there exists a $c \in (0, 1)$ with $f(c)=k$.

Step 3: Evaluate the options.
* "There exists $c \in (0, 1)$ such that $f(c)=12$": $12$ is not between $5$ and $10$. Not guaranteed.
* "There exists $c \in (0, 1)$ such that $f(c)=7$": $7$ is between $5$ and $10$. This is guaranteed by IVT.
"For all $c \in (0, 1)$, $5 < f(c) < 10$": This is not guaranteed. The function could go outside this range and come back, or simply not attain all values. For example, $f(x) = 5+5x$ would satisfy this, but $f(x) = 5+20x(1-x)$ might go above 10. IVT only guarantees existence for values between* endpoints, not that all values are within.
* "There exists $c \in (0, 1)$ such that $f(c)=0$": $0$ is not between $5$ and $10$. Not guaranteed.

The correct option is There exists $c \in (0, 1)$ such that $f(c)=7$."
:::

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9. Extreme Value Theorem (EVT)

The EVT states that a continuous function on a closed, bounded interval must attain both a maximum and a minimum value within that interval.

Quick Example:
Consider $f(x) = x^2$ on the interval $[-1, 2]$. Find the maximum and minimum values.

Step 1: Check continuity and interval type.
$f(x) = x^2$ is continuous on $[-1, 2]$, which is a closed and bounded interval. Thus, EVT applies.

Step 2: Find critical points in the interval.
$f'(x) = 2x$. Setting $f'(x)=0$ gives $x=0$. $x=0$ is in $[-1, 2]$.

Step 3: Evaluate $f(x)$ at critical points and endpoints.

Step 4: Identify minimum and maximum.
The minimum value is $0$ (at $x=0$). The maximum value is $4$ (at $x=2$).

Answer: Minimum value is $0$, maximum value is $4$.

:::question type="MCQ" question="The image of a closed interval under a continuous function is:" options=["Closed interval","Either closed interval or open interval","Open interval","Closed interval or a singleton"] answer="Closed interval or a singleton" hint="Consider the implications of the Extreme Value Theorem and the Intermediate Value Theorem for continuous functions on compact sets. A singleton set is a degenerate closed interval." solution="Step 1: Recall the Extreme Value Theorem (EVT).
If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains an absolute maximum value $M$ and an absolute minimum value $m$ on $[a, b]$.

Step 2: Recall the Intermediate Value Theorem (IVT).
If $f$ is continuous on $[a, b]$, it takes on every value between $f(a)$ and $f(b)$.

Step 3: Combine EVT and IVT.
The EVT guarantees that $f$ attains $m$ and $M$. The IVT then guarantees that $f$ takes on every value between $m$ and $M$. Therefore, the image $f([a, b])$ is the closed interval $[m, M]$.

Step 4: Consider edge cases.
If $f$ is a constant function on $[a, b]$, say $f(x) = k$ for all $x \in [a, b]$, then its minimum $m=k$ and maximum $M=k$. In this case, the image is $[k, k]$, which is a singleton set $\{k\}$. A singleton set can be considered a degenerate closed interval.

Therefore, the image of a closed interval under a continuous function is a closed interval or a singleton.

The correct option is Closed interval or a singleton."
:::

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Advanced Concepts

1. Uniform Continuity

Uniform continuity is a stronger condition than pointwise continuity, concerning the relationship between $\epsilon$ and $\delta$. For a uniformly continuous function, a single $\delta$ works for all points in the domain, given an $\epsilon$.

Quick Example:
Show that $f(x) = 2x+1$ is uniformly continuous on $\mathbb{R}$.

Step 1: Start with $|f(x) - f(y)| < \epsilon$.

Step 2: Simplify the expression.

Step 3: Isolate $|x - y|$.

Step 4: Choose $\delta$. We choose $\delta = \frac{\epsilon}{2}$.
Since $\delta$ depends only on $\epsilon$ (and not on $x$ or $y$), $f(x) = 2x+1$ is uniformly continuous on $\mathbb{R}$.

Answer: Uniformly continuous.

:::question type="MCQ" question="The set on which $f(x) = x^2$ is uniformly continuous is:" options=["$[-1,1]$","$(-1,1)$","$\mathbb{R}$","Nowhere"] answer="$[-1,1]$,$(-1,1)$" type="MSQ" hint="A continuous function on a compact set is uniformly continuous. For non-compact sets, consider the definition. $f(x)=x^2$ is not uniformly continuous on $\mathbb{R}$ because its slope is unbounded." solution="We analyze each option:

Step 1: Recall the Heine-Cantor Theorem.
A function that is continuous on a closed and bounded interval (a compact set) is uniformly continuous on that interval.

Step 2: Analyze $f(x) = x^2$.
This function is continuous on all of $\mathbb{R}$.

Step 3: Evaluate the options based on the definition or theorems.
* $[-1,1]$: This is a closed and bounded interval (compact set). Since $f(x)=x^2$ is continuous on $[-1,1]$, it is uniformly continuous on $[-1,1]$ by the Heine-Cantor Theorem.
* $(-1,1)$: This is an open and bounded interval. While not compact, $f(x)=x^2$ is uniformly continuous on any bounded interval. To show this, for any $\epsilon > 0$, we need to find $\delta$.

If $x, y \in (-1, 1)$, then $|x| < 1$ and $|y| < 1$, so $|x+y| \le |x| + |y| < 1 + 1 = 2$.
Thus, $|f(x) - f(y)| < |x-y| \cdot 2$. If we choose $\delta = \epsilon/2$, then for $|x-y|<\delta$, we have $|f(x)-f(y)| < (\epsilon/2) \cdot 2 = \epsilon$. So, $f(x)=x^2$ is uniformly continuous on $(-1,1)$.
$\mathbb{R}$: This is an unbounded interval. For $f(x)=x^2$ on $\mathbb{R}$, it is not* uniformly continuous. Consider $\epsilon = 1$. We need to find $\delta > 0$ such that if $|x-y| < \delta$, then $|x^2-y^2| < 1$.
Let $x_n = n$ and $y_n = n + \frac{\delta}{2}$. Then $|x_n - y_n| = \frac{\delta}{2} < \delta$.
But

As $n \to \infty$, $n\delta + \frac{\delta^2}{4}$ grows unboundedly, so it will eventually exceed $1$. Thus, no single $\delta$ works for all $x,y \in \mathbb{R}$. Hence, $f(x)=x^2$ is not uniformly continuous on $\mathbb{R}$.
* Nowhere: This is incorrect as it is uniformly continuous on bounded intervals.

Therefore, $f(x)=x^2$ is uniformly continuous on $[-1,1]$ and $(-1,1)$.

The correct options are $[-1,1]$,$(-1,1)$."
:::

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2. Limits at Infinity and Infinite Limits

We examine function behavior as the input grows unboundedly or as the function's output grows unboundedly.

Quick Example:
Evaluate $\lim_{x \to \infty} \frac{2x^2 + 3x - 1}{x^2 - 4x + 5}$.

Step 1: Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$.
>

Step 2: Simplify the expression.
>

Step 3: Apply limit properties. As $x \to \infty$, terms like $\frac{k}{x^n}$ approach $0$.
>

Answer: $2$.

:::question type="MCQ" question="Evaluate $\lim_{x \to -\infty} \frac{\sqrt{x^2+1}}{x}$." options=["$1$","$-1$","0","DoesNotExist"] answer="-1" hint="For limits at infinity involving square roots, factor out $x^2$ from under the root. Remember that $\sqrt{x^2}=|x|$, and for $x \to -\infty$, $|x| = -x$." solution="Step 1: Factor $x^2$ out from under the square root in the numerator.
>

Step 2: Use the property $\sqrt{x^2} = |x|$.
>

Step 3: Since $x \to -\infty$, $x$ is negative, so $|x| = -x$.
>

Step 4: Cancel $x$.
>

Step 5: Evaluate the limit. As $x \to -\infty$, $\frac{1}{x^2} \to 0$.
>

Answer: \boxed{-1}"
:::

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3. Squeeze Theorem

The Squeeze Theorem (also known as the Sandwich Theorem) is useful for finding the limit of a function that is "squeezed" between two other functions whose limits are known and equal.

Quick Example:
Evaluate $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

Step 1: Establish bounds for the oscillating part.
We know that $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ for $x \ne 0$.

Step 2: Multiply by $x^2$. Since $x^2 \ge 0$, the inequalities are preserved.
>

Step 3: Evaluate the limits of the bounding functions as $x \to 0$.
>

>

Step 4: Apply the Squeeze Theorem.
Since $x^2 \sin\left(\frac{1}{x}\right)$ is squeezed between $-x^2$ and $x^2$, and both bounding functions approach $0$ as $x \to 0$, we conclude that $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

Answer: $0$.

:::question type="MCQ" question="Evaluate $\lim_{x \to \infty} \frac{\cos x}{x}$." options=["$1$","0","DoesNotExist","$\infty$"] answer="0" hint="Use the bounds of the cosine function and apply the Squeeze Theorem for limits at infinity." solution="Step 1: Establish bounds for the numerator.
We know that for all real $x$, $-1 \le \cos x \le 1$.

Step 2: Divide the inequality by $x$. Since we are considering $x \to \infty$, $x$ is positive, so the inequalities are preserved.
>

Step 3: Evaluate the limits of the bounding functions as $x \to \infty$.
>

>

Step 4: Apply the Squeeze Theorem.
Since $\frac{\cos x}{x}$ is squeezed between $-\frac{1}{x}$ and $\frac{1}{x}$, and both bounding functions approach $0$ as $x \to \infty$, we conclude that $\lim_{x \to \infty} \frac{\cos x}{x} = 0$.

Answer: \boxed{0}"
:::

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4. L'Hôpital's Rule

L'Hôpital's Rule is a technique for evaluating limits of indeterminate forms, specifically $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It involves taking the derivatives of the numerator and denominator.

Quick Example:
Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$.

Step 1: Check the indeterminate form.
As $x \to 0$, $\sin x \to 0$ and $x \to 0$. This is of the form $\frac{0}{0}$.

Step 2: Apply L'Hôpital's Rule: take derivatives of the numerator and denominator.
$f(x) = \sin x \implies f'(x) = \cos x$
$g(x) = x \implies g'(x) = 1$

>

Step 3: Evaluate the new limit.
>

Answer: $1$.

:::question type="MCQ" question="Evaluate $\lim_{x \to 1} \frac{\ln x}{x-1}$." options=["$0$","$1$","$e$","DoesNotExist"] answer="1" hint="Check if the limit is an indeterminate form. If it is, apply L'Hôpital's Rule by differentiating the numerator and denominator." solution="Step 1: Check the indeterminate form.
As $x \to 1$, $\ln x \to \ln 1 = 0$.
As $x \to 1$, $x-1 \to 1-1 = 0$.
The limit is of the indeterminate form $\frac{0}{0}$.

Step 2: Apply L'Hôpital's Rule.
Let $f(x) = \ln x$ and $g(x) = x-1$.
Then $f'(x) = \frac{1}{x}$ and $g'(x) = 1$.

>

Step 3: Evaluate the new limit.
>

Answer: \boxed{1}"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let

For $f(x)$ to be continuous at $x=0$, what must be the values of $a$ and $b$?" options=["$a=b$","Values of $a$ and $b$ can be any real numbers","There is no such $a$ and $b$","$a=b$ and $a \ne 0$"] answer="$a=b$" hint="For continuity at $x=0$, the left-hand limit, right-hand limit, and $f(0)$ must all be equal. Use the standard limit $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$." solution="Step 1: For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

Step 2: Evaluate the left-hand limit.

Using the standard limit $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$, we have:

Step 3: Evaluate the right-hand limit.

Step 4: Evaluate $f(0)$.
Using the second case ($x \ge 0$):

Step 5: Equate the results for continuity.
For continuity, $a = b$. There is no restriction that $a \ne 0$. If $a=0$, then $b=0$, and $f(x)=0$ for all $x$, which is continuous.

The correct option is $a=b$."
:::

:::question type="NAT" question="If $f(x) = \frac{x^2-1}{x-1}$ for $x \ne 1$, what value must $f(1)$ be defined as to make $f(x)$ continuous at $x=1$?" answer="2" hint="For removable discontinuity, $f(1)$ must be equal to $\lim_{x \to 1} f(x)$." solution="Step 1: For $f(x)$ to be continuous at $x=1$, we must define $f(1)$ such that $f(1) = \lim_{x \to 1} f(x)$.

Step 2: Evaluate the limit $\lim_{x \to 1} f(x)$.

This is of the form $\frac{0}{0}$. We can factor the numerator.

For $x \ne 1$, we can cancel $(x-1)$.

Step 3: Define $f(1)$ to be this limit.
Therefore, $f(1)$ must be $2$ for the function to be continuous at $x=1$.

The answer is 2."
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:::question type="MCQ" question="Let $f(x) = \frac{1}{x-2}$. Which of the following statements is true about its uniform continuity?" options=["Uniformly continuous on $(2, \infty)$","Uniformly continuous on $[3, 5]$","Uniformly continuous on $(0, 3)$","Uniformly continuous on $\mathbb{R} \setminus \{2\}$"] answer="Uniformly continuous on $[3, 5]$" hint="Recall that a continuous function on a closed and bounded interval is uniformly continuous. Check other intervals for unboundedness of the derivative or the function itself." solution="Step 1: Analyze the continuity of $f(x) = \frac{1}{x-2}$.
This function is continuous on its domain, which is $\mathbb{R} \setminus \{2\}$.

Step 2: Evaluate each option for uniform continuity.
* Uniformly continuous on $(2, \infty)$: This interval is unbounded. Consider $x_n = 2 + \frac{1}{n}$ and $y_n = 2 + \frac{1}{n+1}$. $|x_n - y_n|$ can be made arbitrarily small. However, $f(x_n) = n$ and $f(y_n) = n+1$. The difference $|f(x_n) - f(y_n)| = 1$, which does not go to 0. Alternatively, the derivative $f'(x) = -\frac{1}{(x-2)^2}$ is unbounded as $x \to 2^+$. Thus, it is not uniformly continuous on $(2, \infty)$.
* Uniformly continuous on $[3, 5]$: This is a closed and bounded interval (a compact set). Since $f(x) = \frac{1}{x-2}$ is continuous on $[3, 5]$ (as $2 \notin [3, 5]$), by the Heine-Cantor Theorem, $f(x)$ is uniformly continuous on $[3, 5]$.
* Uniformly continuous on $(0, 3)$: This interval contains $x=2$, where the function is discontinuous. A function cannot be uniformly continuous on an interval where it is not even continuous.
* Uniformly continuous on $\mathbb{R} \setminus \{2\}$: This is an unbounded interval and also contains points where the derivative is unbounded (near $x=2$). It is not uniformly continuous.

The correct option is Uniformly continuous on $[3, 5]$."
:::

:::question type="MCQ" question="Let

Which of the following statements is correct?" options=["$f(x)$ is continuous on $[0,1]$","$f(x)$ is continuous at $x=1/2$","$f(x)$ is discontinuous at $x=1/2$","$f(x)$ is continuous on $(1/2,1)$"] answer="$f(x)$ is continuous at $x=1/2$" type="MSQ" hint="Analyze the definition of the function across different intervals. The intervals are of the form $[1/(r+1), 1/r)$. For $r=1$, $x \in [1/2, 1)$. For $r=2$, $x \in [1/3, 1/2)$. For $r=3$, $x \in [1/4, 1/3)$." solution="We analyze the function $f(x)$ and its continuity.

The definition partitions the interval $(0, 1)$ into disjoint intervals:
For $r=1$, $x \in [1/2, 1)$, $f(x) = (-1)^1 = -1$.
For $r=2$, $x \in [1/3, 1/2)$, $f(x) = (-1)^2 = 1$.
For $r=3$, $x \in [1/4, 1/3)$, $f(x) = (-1)^3 = -1$.
And so on.

Option 1: $f(x)$ is continuous on $[0,1]$
This is false. Consider $x=1/2$.
$\lim_{x \to (1/2)^-} f(x) = \lim_{x \to (1/2)^-} (-1)^2 = 1$ (from $r=2$ interval)
$\lim_{x \to (1/2)^+} f(x) = \lim_{x \to (1/2)^+} (-1)^1 = -1$ (from $r=1$ interval)
Since the one-sided limits are not equal, $f(x)$ is discontinuous at $x=1/2$.
Also, as $x \to 0^+$, $f(x)$ oscillates between $1$ and $-1$ (e.g., $f(1/2k) = 1$ and $f(1/(2k+1)) = -1$). So $\lim_{x \to 0^+} f(x)$ does not exist, and $f(x)$ is not continuous at $x=0$.
Therefore, $f(x)$ is not continuous on $[0,1]$.

Option 2: $f(x)$ is continuous at $x=1/2$
This is false, as shown above. The left-hand limit is $1$ and the right-hand limit is $-1$.

Option 3: $f(x)$ is discontinuous at $x=1/2$
This is true, as shown above.

Option 4: $f(x)$ is continuous on $(1/2,1)$
For $x \in (1/2, 1)$, $x$ falls into the interval $[1/2, 1)$ (for $r=1$). In this interval, $f(x) = (-1)^1 = -1$.
The function is a constant $-1$ on the open interval $(1/2, 1)$. Constant functions are continuous.
Therefore, $f(x)$ is continuous on $(1/2,1)$.

The correct options are $f(x)$ is discontinuous at $x=1/2$, $f(x)$ is continuous on $(1/2,1)$."
:::

:::question type="MCQ" question="Let $f: [a, b] \to (a, b)$ be a continuous function. Which of the following statements is true?" options=["$f$ can be surjective","The range of $f$ is an open interval","The range of $f$ is a closed interval","The range of $f$ is a singleton set"] answer="The range of $f$ is a closed interval" hint="Apply the Extreme Value Theorem and the Intermediate Value Theorem. The image of a closed interval under a continuous function is always a closed interval (possibly a singleton). Consider if it can be surjective onto an open interval." solution="Step 1: Apply the Extreme Value Theorem (EVT).
Since $f$ is a continuous function on a closed and bounded interval $[a, b]$, the EVT guarantees that $f$ attains an absolute maximum value $M$ and an absolute minimum value $m$ on $[a, b]$.

Step 2: Apply the Intermediate Value Theorem (IVT).
Since $f$ is continuous on $[a, b]$, and it attains all values between $m$ and $M$, the image of $f$ on $[a, b]$ is the closed interval $[m, M]$.
So, the range of $f$ is $[m, M]$.

Step 3: Analyze the given codomain $(a, b)$.
The range of $f$ is $[m, M]$. We are given that $f: [a, b] \to (a, b)$, which means $m, M \in (a, b)$.
So, $a < m \le M < b$.
Since the range is $[m, M]$, it is a closed interval.

Step 4: Consider the options.
"$f$ can be surjective": For $f$ to be surjective onto $(a,b)$, its range must be $(a,b)$. However, the range of $f$ is $[m,M]$, which is a closed interval. A closed interval cannot be equal to an open interval. Specifically, $[m,M]$ must contain its endpoints $m$ and $M$. If $f$ were surjective onto $(a,b)$, it would need to attain values arbitrarily close to $a$ and $b$, but not $a$ or $b$ themselves. But by EVT, it must attain a min $m$ and a max $M$, and these must be within* $(a,b)$. If $f$ were surjective onto $(a,b)$, then $m$ would have to be $a$ and $M$ would have to be $b$, which contradicts $m,M \in (a,b)$. Hence $f$ cannot be surjective.
* "The range of $f$ is an open interval": This contradicts the conclusion from EVT and IVT that the range is a closed interval $[m,M]$.
* "The range of $f$ is a closed interval": This is consistent with EVT and IVT.
"The range of $f$ is a singleton set": This is a special case of a closed interval (e.g., if $f(x)=c$ for all $x$). If $f(x)$ is a constant function, its range is a singleton, which is a closed interval. However, the question asks what is true. The range is always* a closed interval (which includes the singleton case).

The most accurate and general statement is that the range of $f$ is a closed interval.

The correct option is The range of $f$ is a closed interval."
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Summary

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What's Next?

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Part 2: Differentiation and its Theorems

Differentiation forms a foundational pillar of calculus, enabling us to quantify rates of change and analyze the local behavior of functions. We explore the rigorous definition of differentiability, fundamental rules, and critical theorems such as Rolle's, Lagrange's, and Cauchy's Mean Value Theorems, which are indispensable for understanding function properties and solving advanced problems in real analysis.

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Core Concepts

1. Definition of Differentiability

A function $f: (a,b) \to \mathbb{R}$ is differentiable at a point $c \in (a,b)$ if the limit of the difference quotient exists. This limit, if it exists, is denoted by $f'(c)$.

For the limit to exist, the left-hand derivative (LHD) and the right-hand derivative (RHD) must be equal.

Relationship with Continuity: If a function is differentiable at a point, it must be continuous at that point. However, the converse is not true; continuity does not guarantee differentiability.

Quick Example: Determine if $f(x) = |x|$ is differentiable at $x=0$.

Step 1: Calculate the Left-Hand Derivative (LHD) at $x=0$.

Step 2: Calculate the Right-Hand Derivative (RHD) at $x=0$.

Step 3: Compare LHD and RHD.
Since $LHD \neq RHD$ ($-1 \neq 1$), $f(x) = |x|$ is not differentiable at $x=0$.

:::question type="MCQ" question="Let $f(x) = \begin{cases} x^2 \sin(1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$. Which of the following statements is true about $f(x)$ at $x=0$?" options=["$f(x)$ is continuous but not differentiable at $x=0$.","$f(x)$ is differentiable but not continuous at $x=0$.","$f(x)$ is both continuous and differentiable at $x=0$.","$f(x)$ is neither continuous nor differentiable at $x=0$."] answer="$f(x)$ is both continuous and differentiable at $x=0$." hint="First check continuity, then differentiability using the definition of the derivative." solution="Step 1: Check Continuity at $x=0$.
We need to check if $\lim_{x \to 0} f(x) = f(0)$.
We have $f(0) = 0$.
Consider the limit:

Since $-1 \le \sin(1/x) \le 1$, we have $-x^2 \le x^2 \sin(1/x) \le x^2$.
As $x \to 0$, $x^2 \to 0$. By the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin(1/x) = 0$.
Since $\lim_{x \to 0} f(x) = f(0)$, $f(x)$ is continuous at $x=0$.

Step 2: Check Differentiability at $x=0$.
We use the definition of the derivative:

Again, by the Squeeze Theorem, since $-|h| \le h \sin(1/h) \le |h|$ and $\lim_{h \to 0} |h| = 0$, we have $\lim_{h \to 0} h \sin(1/h) = 0$.
Since the limit exists, $f(x)$ is differentiable at $x=0$, and $f'(0)=0$.
Therefore, $f(x)$ is both continuous and differentiable at $x=0$.
Answer: $\boxed{\text{f(x) is both continuous and differentiable at } x=0}$
"
:::

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2. Rules of Differentiation

We recall the fundamental rules for differentiating combinations of functions. These rules are crucial for computing derivatives efficiently.

Quick Example: Differentiate $y = \sin(e^{x^2})$.

Step 1: Identify the composite functions.
Let $y = \sin(u)$, where $u = e^v$, and $v = x^2$.

Step 2: Differentiate each layer.

Step 3: Apply the Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

Answer: $\boxed{2xe^{x^2}\cos(e^{x^2})}$

:::question type="MCQ" question="If $y = \ln(\sec x + \tan x)$, find $\frac{dy}{dx}$." options=["$\sec x$","$\tan x$","$\sin x$","$\cos x$"] answer="$\sec x$" hint="Use the chain rule and the derivatives of $\ln u$, $\sec x$, and $\tan x$." solution="Step 1: Apply the Chain Rule.
Let $u = \sec x + \tan x$. Then $y = \ln u$.

Step 2: Differentiate $u$ with respect to $x$.
Recall that $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$.

Step 3: Substitute back into the chain rule expression.

Step 4: Simplify the expression.

Answer: $\boxed{\sec x}$
"
:::

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3. Differentiability of Absolute Value and Piecewise Functions

Functions involving absolute values or defined piecewise often exhibit points where differentiability fails, typically at "corners" or "cusps." These points occur where the argument of the absolute value function changes sign or where the definition of the piecewise function switches.

We must examine the LHD and RHD at such critical points to determine differentiability. For a sum of absolute value functions, the critical points are where any of the terms inside the absolute value become zero.

Quick Example: For $f(x) = |x| + |x-1|$, find the points where $f(x)$ is not differentiable.

Step 1: Identify critical points.
The critical points are $x=0$ (from $|x|$) and $x=1$ (from $|x-1|$).

Step 2: Define $f(x)$ piecewise for different intervals.
* For $x < 0$: $f(x) = -x + (-(x-1)) = -x - x + 1 = -2x + 1$
* For $0 \le x < 1$: $f(x) = x + (-(x-1)) = x - x + 1 = 1$
* For $x \ge 1$: $f(x) = x + (x-1) = 2x - 1$

So,

Step 3: Calculate derivatives for each segment.

Step 4: Check differentiability at critical points $x=0$ and $x=1$.

At $x=0$:

Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=0$.

At $x=1$:

Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=1$.

Answer: $f(x)$ is not differentiable at $x=0$ and $x=1$.

:::question type="MCQ" question="Let $f(x) = |x| + |x-1| + |x+1|$ be a function defined on $\mathbb{R}$. Then $f'(x)$ is:" options=["differentiable for all $x \in \mathbb{R}$","differentiable for all $x \in \mathbb{R}$ other than $x = -1, 0, 1$","differentiable only for $x = -1, 0, 1$","not differentiable at any real point"] answer="differentiable for all $x \in \mathbb{R}$ other than $x = -1, 0, 1$" hint="Identify the critical points where the arguments of the absolute value functions become zero. Analyze the function's piecewise definition around these points." solution="Step 1: Identify critical points.
The critical points are where the terms inside the absolute values become zero: $x=0$, $x=1$, and $x=-1$. These are the potential points of non-differentiability.

Step 2: Define $f(x)$ piecewise.
* For $x < -1$: $f(x) = -x + (-(x-1)) + (-(x+1)) = -x - x + 1 - x - 1 = -3x$
* For $-1 \le x < 0$: $f(x) = -x + (-(x-1)) + (x+1) = -x - x + 1 + x + 1 = -x + 2$
* For $0 \le x < 1$: $f(x) = x + (-(x-1)) + (x+1) = x - x + 1 + x + 1 = x + 2$
* For $x \ge 1$: $f(x) = x + (x-1) + (x+1) = 3x$

So,

Step 3: Calculate derivatives for each segment.

Step 4: Check differentiability at critical points $x=-1, 0, 1$.
* At $x=-1$: $LHD = -3$, $RHD = -1$. Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=-1$.
* At $x=0$: $LHD = -1$, $RHD = 1$. Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=0$.
* At $x=1$: $LHD = 1$, $RHD = 3$. Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x=1$.

For all other points (i.e., $x \in \mathbb{R} \setminus \{-1, 0, 1\}$), the derivative exists and is constant in the respective intervals.
Therefore, $f(x)$ is differentiable for all $x \in \mathbb{R}$ other than $x = -1, 0, 1$."
:::

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4. Rolle's Theorem

Rolle's Theorem provides a condition for the existence of a point where the derivative of a function is zero. It is a special case of the Mean Value Theorem.

Geometric Interpretation: If a continuous and differentiable curve starts and ends at the same height, there must be at least one point between the endpoints where the tangent line is horizontal.

Quick Example: Verify Rolle's Theorem for $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$ and find the value of $c$.

Step 1: Check continuity.
$f(x)$ is a polynomial, hence continuous on $[1, 3]$.

Step 2: Check differentiability.
$f(x)$ is a polynomial, hence differentiable on $(1, 3)$. $f'(x) = 2x - 4$.

Step 3: Check $f(a) = f(b)$.

Since $f(1) = f(3) = 0$, all conditions are satisfied.

Step 4: Find $c \in (1, 3)$ such that $f'(c) = 0$.

The value $c=2$ lies in the interval $(1, 3)$.

Answer: Rolle's Theorem applies, and $c=2$.

:::question type="MCQ" question="Which of the following functions satisfies Rolle's theorem on the given interval?" options=["$f(x) = |x|$ in $[-1, 1]$","$f(x) = \frac{1}{x}$ in $[-1, 1]$","$f(x) = \sin x$ in $[0, 2\pi]$","$f(x) = x^2$ in $[-1, 0]$"] answer="$f(x) = \sin x$ in $[0, 2\pi]$" hint="Check all three conditions (continuity, differentiability, $f(a)=f(b)$) for each option." solution="Let's check each option:

Option 1: $f(x) = |x|$ in $[-1, 1]$

Continuous on $[-1, 1]$: Yes.

Differentiable on $(-1, 1)$: No, $f(x)$ is not differentiable at $x=0 \in (-1, 1)$.

Since differentiability fails, Rolle's Theorem is not applicable.

Option 2: $f(x) = \frac{1}{x}$ in $[-1, 1]$

Continuous on $[-1, 1]$: No, $f(x)$ is not defined at $x=0 \in [-1, 1]$, hence not continuous.

Since continuity fails, Rolle's Theorem is not applicable.

Option 3: $f(x) = \sin x$ in $[0, 2\pi]$

Continuous on $[0, 2\pi]$: Yes, $\sin x$ is continuous everywhere.

Differentiable on $(0, 2\pi)$: Yes, $\sin x$ is differentiable everywhere ($f'(x) = \cos x$).

$f(0) = \sin(0) = 0$. $f(2\pi) = \sin(2\pi) = 0$. So, $f(0) = f(2\pi)$.

All conditions are satisfied. Thus, Rolle's Theorem is applicable for $f(x) = \sin x$ in $[0, 2\pi]$.

Option 4: $f(x) = x^2$ in $[-1, 0]$

Continuous on $[-1, 0]$: Yes.

Differentiable on $(-1, 0)$: Yes.

$f(-1) = (-1)^2 = 1$. $f(0) = (0)^2 = 0$. Since $f(-1) \neq f(0)$, the third condition fails.

Rolle's Theorem is not applicable.

The only function that satisfies Rolle's Theorem is $f(x) = \sin x$ in $[0, 2\pi]$."
:::

:::question type="MCQ" question="If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$, then $\int_1^2 f'(x) dx$ is equal to" options=["$1.4$","$2.0$","$3.1$","$0$"] answer="$0$" hint="Recall the Fundamental Theorem of Calculus and the third condition of Rolle's Theorem." solution="Step 1: Apply the Fundamental Theorem of Calculus.
The definite integral of a derivative $f'(x)$ from $a$ to $b$ is given by:

For the given problem, $a=1$ and $b=2$.

Step 2: Apply Rolle's Theorem conditions.
The problem states that $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$.
One of the conditions for Rolle's Theorem is $f(a) = f(b)$.
Therefore, $f(1) = f(2)$.

Step 3: Substitute the condition into the integral result.
Since $f(1) = f(2)$, we have:

The value of the integral is $0$."
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5. Lagrange's Mean Value Theorem (LMVT)

Lagrange's Mean Value Theorem generalizes Rolle's Theorem by relaxing the condition $f(a)=f(b)$.

Geometric Interpretation: There is at least one point $c$ in the interval $(a,b)$ where the tangent line to the curve $y=f(x)$ is parallel to the secant line connecting the points $(a, f(a))$ and $(b, f(b))$.

Relationship to Rolle's Theorem: If $f(a) = f(b)$, then $\frac{f(b) - f(a)}{b - a} = \frac{0}{b-a} = 0$, which implies $f'(c) = 0$. Thus, Rolle's Theorem is a special case of LMVT.

Quick Example: Verify LMVT for $f(x) = x^3 - 5x^2 - 3x$ on the interval $[1, 3]$ and find the value of $c$.

Step 1: Check continuity.
$f(x)$ is a polynomial, hence continuous on $[1, 3]$.

Step 2: Check differentiability.
$f(x)$ is a polynomial, hence differentiable on $(1, 3)$. $f'(x) = 3x^2 - 10x - 3$.

Step 3: Calculate $f(a)$, $f(b)$, and $\frac{f(b) - f(a)}{b - a}$.

Step 4: Find $c \in (1, 3)$ such that $f'(c) = -10$.

We solve the quadratic equation for $c$:




Two possible values for $c$:


Only $c_1 = \frac{7}{3}$ lies in the open interval $(1, 3)$ (since $1 < \frac{7}{3} < 3$).

Answer: LMVT applies, and $c = \frac{7}{3}$.

:::question type="MCQ" question="For $f(x) = x^2 - 2x + 1$ on the interval $[0, 3]$, find the value of $c$ that satisfies the Lagrange's Mean Value Theorem." options=["$1$","$1.5$","$2$","$2.5$"] answer="$1.5$" hint="Verify continuity and differentiability, then set $f'(c)$ equal to the slope of the secant line." solution="Step 1: Check conditions.
$f(x) = x^2 - 2x + 1$ is a polynomial, so it is continuous on $[0, 3]$ and differentiable on $(0, 3)$.

Step 2: Calculate $f(a)$, $f(b)$, and the slope of the secant line.
$a=0, b=3$.

Slope of the secant line:

Step 3: Find $f'(x)$ and set $f'(c)$ equal to the slope.

Set $f'(c) = 1$:



The value $c=1.5$ lies in the interval $(0, 3)$.
"
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6. Cauchy's Mean Value Theorem (CMVT)

Cauchy's Mean Value Theorem (also known as the Extended Mean Value Theorem) is a generalization of LMVT involving two functions.

Relationship to LMVT: If $g(x) = x$, then $g'(x) = 1$ (which is never zero), and the theorem reduces to LMVT: $\frac{f'(c)}{1} = \frac{f(b) - f(a)}{b - a}$.

Quick Example: Verify CMVT for $f(x) = x^2$ and $g(x) = x^3$ on the interval $[1, 2]$ and find $c$.

Step 1: Check continuity and differentiability.
Both $f(x) = x^2$ and $g(x) = x^3$ are polynomials, hence continuous on $[1, 2]$ and differentiable on $(1, 2)$.

Step 2: Check $g'(x) \neq 0$ on $(1, 2)$.

For $x \in (1, 2)$, $3x^2 > 0$, so $g'(x) \neq 0$. All conditions are satisfied.

Step 3: Calculate $f(a), f(b), g(a), g(b)$.

Step 4: Calculate $f'(x)$ and $g'(x)$.

Step 5: Find $c \in (1, 2)$ such that $\frac{f'(c)}{g'(c)} = \frac{3}{7}$.

The value $c = \frac{14}{9}$ lies in the interval $(1, 2)$ (since $1 < \frac{14}{9} \approx 1.55 < 2$).

Answer: CMVT applies, and $c = \frac{14}{9}$.

:::question type="MCQ" question="Given $f(x) = \sin x$ and $g(x) = \cos x$ on the interval $[0, \pi/2]$, find the value of $c$ that satisfies Cauchy's Mean Value Theorem." options=["$\pi/6$","$\pi/4$","$\pi/3$","$\pi/2$"] answer="$\pi/4$" hint="Check conditions, then equate the ratio of derivatives to the ratio of function differences." solution="Step 1: Check conditions.
Both $f(x) = \sin x$ and $g(x) = \cos x$ are continuous on $[0, \pi/2]$ and differentiable on $(0, \pi/2)$.

On $(0, \pi/2)$, $\sin x > 0$, so $g'(x) \neq 0$. All conditions are met.

Step 2: Calculate $f(a), f(b), g(a), g(b)$.
$a=0, b=\pi/2$.

Step 3: Calculate $f'(x)$ and $g'(x)$.

Step 4: Find $c \in (0, \pi/2)$ such that $\frac{f'(c)}{g'(c)} = -1$.

For $c \in (0, \pi/2)$, $\cot c = 1$ implies $c = \pi/4$.
The value $c = \pi/4$ lies in the interval $(0, \pi/2)$.
"
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Advanced Applications

7. Differentiability from Lipschitz-like Conditions

A function satisfying a specific Lipschitz-like condition can reveal important properties about its differentiability. Consider a condition of the form $|f(x)-f(y)| \leq K|x-y|^n$.

Quick Example: Let $f(x)$ be a real-valued function such that $|f(x)-f(y)| \leq (x-y)^3$ for all $x,y \in \mathbb{R}$. Show that $f(x)$ is a constant function.

Step 1: Consider the definition of the derivative at an arbitrary point $x_0$.
>

Step 2: Apply the given condition to the difference quotient.
>

Step 3: Evaluate the limit as $x \to x_0$.
>

>
Since absolute values are non-negative, this implies $|f'(x_0)| = 0$, so $f'(x_0) = 0$.

Step 4: Conclude the nature of the function.
Since $f'(x) = 0$ for all $x \in \mathbb{R}$, $f(x)$ must be a constant function.

Answer: $f(x)$ is a constant function.

:::question type="MCQ" question="Let $f(x)$ be a real valued function defined for all $x \in \mathbb{R}$, such that $|f(x)-f(y)| \leq (x-y)^2 \forall x,y \in \mathbb{R}$. Then" options=["$f(x)$ is nowhere differentiable","$f(x)$ is a constant function","$f(x)$ is strictly increasing function in the interval $[0,1]$","$f(x)$ is strictly increasing function for all $x \in \mathbb{R}$"] answer="$f(x)$ is a constant function" hint="Use the definition of the derivative and the given inequality to show that $f'(x) = 0$ everywhere." solution="Step 1: Consider the derivative at an arbitrary point $x_0$.
The derivative of $f$ at $x_0$ is given by:
>

Step 2: Use the given inequality.
We are given $|f(x)-f(y)| \leq (x-y)^2$. Let $y=x_0$.
>

For $x \neq x_0$, we can divide by $|x-x_0|$:
>

Step 3: Take the limit as $x \to x_0$.
>

>
Since $|f'(x_0)|$ must be non-negative, the only possibility is $|f'(x_0)| = 0$, which implies $f'(x_0) = 0$.

Step 4: Conclude the nature of $f(x)$.
Since $f'(x) = 0$ for all $x \in \mathbb{R}$, the function $f(x)$ must be a constant function.
Answer: \boxed{f(x) \text{ is a constant function}}
"
:::

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8. Monotonicity and Extrema

The first and second derivatives are powerful tools for analyzing the behavior of functions, including their monotonicity (increasing or decreasing) and identifying local extrema (maxima and minima).

Quick Example: Find the intervals of increase/decrease and local extrema for $f(x) = x^3 - 3x^2 + 2$.

Step 1: Find the first derivative and critical points.
>

Set $f'(x) = 0$ to find critical points: $3x(x-2) = 0 \implies x=0$ or $x=2$.

Step 2: Determine intervals of increase/decrease using a sign chart for $f'(x)$.
* For $x < 0$: Choose $x=-1$, $f'(-1) = 3(-1)(-1-2) = (-3)(-3) = 9 > 0$. ($f$ is increasing)
* For $0 < x < 2$: Choose $x=1$, $f'(1) = 3(1)(1-2) = (3)(-1) = -3 < 0$. ($f$ is decreasing)
* For $x > 2$: Choose $x=3$, $f'(3) = 3(3)(3-2) = (9)(1) = 9 > 0$. ($f$ is increasing)

Step 3: Identify local extrema using the First Derivative Test.
* At $x=0$, $f'(x)$ changes from positive to negative. Local maximum at $x=0$.
$f(0) = (0)^3 - 3(0)^2 + 2 = 2$. Local maximum is $(0, 2)$.
* At $x=2$, $f'(x)$ changes from negative to positive. Local minimum at $x=2$.
$f(2) = (2)^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$. Local minimum is $(2, -2)$.

Answer:
* Increasing on $(-\infty, 0) \cup (2, \infty)$.
* Decreasing on $(0, 2)$.
* Local maximum at $(0, 2)$.
* Local minimum at $(2, -2)$.

:::question type="NAT" question="Find the local maximum value of the function $f(x) = x^3 - 6x^2 + 9x + 1$." answer="5" hint="Find critical points using the first derivative, then use the first or second derivative test to classify them." solution="Step 1: Find the first derivative.
>

Step 2: Find critical points by setting $f'(x) = 0$.
>

Divide by 3:
>
Factor the quadratic:
>
Critical points are $x=1$ and $x=3$.

Step 3: Use the Second Derivative Test to classify critical points.
Find the second derivative:
>

Evaluate $f''(x)$ at each critical point:
* At $x=1$:
>
Since $f''(1) < 0$, there is a local maximum at $x=1$.
* At $x=3$:
>
Since $f''(3) > 0$, there is a local minimum at $x=3$.

Step 4: Calculate the local maximum value.
The local maximum occurs at $x=1$.
>

Answer: \boxed{5}
"
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9. Concavity and Inflection Points

The second derivative also describes the concavity of a function's graph and helps identify inflection points.

Quick Example: Find the intervals of concavity and inflection points for $f(x) = x^4 - 4x^3$.

Step 1: Find the first and second derivatives.
>

>

Step 2: Find potential inflection points by setting $f''(x) = 0$.
$12x(x-2) = 0 \implies x=0$ or $x=2$.

Step 3: Determine intervals of concavity using a sign chart for $f''(x)$.
* For $x < 0$: Choose $x=-1$, $f''(-1) = 12(-1)(-1-2) = (-12)(-3) = 36 > 0$. (Concave up)
* For $0 < x < 2$: Choose $x=1$, $f''(1) = 12(1)(1-2) = (12)(-1) = -12 < 0$. (Concave down)
* For $x > 2$: Choose $x=3$, $f''(3) = 12(3)(3-2) = (36)(1) = 36 > 0$. (Concave up)

Step 4: Identify inflection points.
* At $x=0$, $f''(x)$ changes from positive to negative. Inflection point at $x=0$.
$f(0) = (0)^4 - 4(0)^3 = 0$. Point is $(0, 0)$.
* At $x=2$, $f''(x)$ changes from negative to positive. Inflection point at $x=2$.
$f(2) = (2)^4 - 4(2)^3 = 16 - 4(8) = 16 - 32 = -16$. Point is $(2, -16)$.

Answer:
* Concave up on $(-\infty, 0) \cup (2, \infty)$.
* Concave down on $(0, 2)$.
* Inflection points at $(0, 0)$ and $(2, -16)$.

:::question type="MCQ" question="For the function $f(x) = x^4 - 6x^2$, identify the intervals where the function is concave down." options=["$(-\infty, -1) \cup (1, \infty)$","$(-1, 1)$","$(-\infty, 0) \cup (0, \infty)$","$(-\infty, -2) \cup (2, \infty)$"] answer="$(-1, 1)$" hint="Compute the second derivative and find where it is negative." solution="Step 1: Compute the first derivative.
>

Step 2: Compute the second derivative.
>

Step 3: Find points where $f''(x) = 0$ or is undefined.
>

>
>
>
So, $x=1$ and $x=-1$ are potential inflection points.

Step 4: Determine the sign of $f''(x)$ in intervals defined by these points.
* For $x < -1$: Choose $x=-2$. $f''(-2) = 12(-2)^2 - 12 = 12(4) - 12 = 48 - 12 = 36 > 0$. (Concave up)
* For $-1 < x < 1$: Choose $x=0$. $f''(0) = 12(0)^2 - 12 = -12 < 0$. (Concave down)
* For $x > 1$: Choose $x=2$. $f''(2) = 12(2)^2 - 12 = 12(4) - 12 = 48 - 12 = 36 > 0$. (Concave up)

Step 5: Identify intervals of concavity.
The function is concave down when $f''(x) < 0$. This occurs on the interval $(-1, 1)$.
Answer: \boxed{(-1, 1)}
"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $f(x) = x^{2/3}$ on the interval $[-1, 1]$. Which of the following statements is true?" options=["Rolle's Theorem is applicable to $f(x)$ on $[-1, 1]$.","Lagrange's Mean Value Theorem is applicable to $f(x)$ on $[-1, 1]$.","$f(x)$ is continuous but not differentiable on $(-1, 1)$.","There exists $c \in (-1, 1)$ such that $f'(c) = 0$."] answer="$f(x)$ is continuous but not differentiable on $(-1, 1)$." hint="Check the conditions for continuity and differentiability, especially at $x=0$." solution="Step 1: Check continuity.
$f(x) = x^{2/3} = (x^2)^{1/3}$ is a power function. It is continuous for all real $x$. So, it is continuous on $[-1, 1]$.

Step 2: Check differentiability.
>

This derivative is defined for all $x \neq 0$. At $x=0$, $f'(x)$ is undefined (the denominator becomes zero).
Since $0 \in (-1, 1)$, $f(x)$ is not differentiable on $(-1, 1)$.

Step 3: Evaluate applicability of theorems.
* Rolle's Theorem: Requires differentiability on $(-1, 1)$, which fails at $x=0$. Also, $f(-1) = (-1)^{2/3} = 1$ and $f(1) = (1)^{2/3} = 1$, so $f(a)=f(b)$ condition is met. However, differentiability fails. So, Rolle's Theorem is not applicable.
* Lagrange's Mean Value Theorem: Requires differentiability on $(-1, 1)$, which fails at $x=0$. So, LMVT is not applicable.

Step 4: Evaluate the last option.
If LMVT is not applicable, there is no guarantee for such a $c$. Indeed, $f'(c) = \frac{2}{3\sqrt[3]{c}}$ can never be $0$.

Thus, $f(x)$ is continuous but not differentiable on $(-1, 1)$ is the true statement.
Answer: \boxed{f(x) \text{ is continuous but not differentiable on } (-1, 1)}
"
:::

:::question type="NAT" question="If $f(x) = \frac{x}{2} + \sin x$, find the value of $c \in (0, \pi)$ that satisfies Lagrange's Mean Value Theorem." answer="1.5708" hint="First, ensure the conditions for LMVT are met. Then, calculate $f(0)$, $f(\pi)$, and $f'(x)$. Set $f'(c)$ equal to $\frac{f(\pi) - f(0)}{\pi - 0}$ and solve for $c$ in radians." solution="Step 1: Check LMVT conditions.
$f(x) = \frac{x}{2} + \sin x$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$ (since $x/2$ and $\sin x$ are continuous and differentiable everywhere).

Step 2: Calculate $f(0)$ and $f(\pi)$.
>

>

Step 3: Calculate the slope of the secant line.
>

Step 4: Find $f'(x)$.
>

Step 5: Set $f'(c)$ equal to the slope and solve for $c$.
>

>
For $c \in (0, \pi)$, the only value for which $\cos c = 0$ is $c = \frac{\pi}{2}$.
In decimal form, $c \approx 1.570796...$
Rounding to 4 decimal places, $c = 1.5708$.
Answer: \boxed{1.5708}
"
:::

:::question type="MCQ" question="Let $f(x) = \begin{cases} x^3 & \text{if } x \le 1 \\ ax+b & \text{if } x > 1 \end{cases}$. If $f(x)$ is differentiable at $x=1$, what are the values of $a$ and $b$?" options=["$a=3, b=-2$","$a=1, b=0$","$a=3, b=1$","$a=-1, b=2$"] answer="$a=3, b=-2$" hint="For differentiability, the function must first be continuous. Use continuity and equality of LHD/RHD to form equations for $a$ and $b$." solution="Step 1: For $f(x)$ to be differentiable at $x=1$, it must first be continuous at $x=1$.
Continuity at $x=1$ requires $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
>

>
>
So, for continuity:
>

Step 2: For $f(x)$ to be differentiable at $x=1$, LHD must equal RHD.
First, find the derivative of each piece:
>

Calculate LHD at $x=1$:
>
Calculate RHD at $x=1$:
>
For differentiability, $LHD = RHD$:
>

Step 3: Substitute $a=3$ into equation $(*)$ to find $b$.
>

>
>
Thus, $a=3$ and $b=-2$.
Answer: \boxed{a=3, b=-2}
"
:::

:::question type="MSQ" question="Let $f(x) = (x-1)(x-2)(x-3)$. On which of the following intervals does Rolle's Theorem apply to $f(x)$?" options=["$[1, 2]$","$[2, 3]$","$[1, 3]$","$[0, 1]$"] answer="$[1, 2]$,$[2, 3]$,$[1, 3]$" hint="Rolle's Theorem requires continuity on $[a,b]$, differentiability on $(a,b)$, and $f(a)=f(b)$. Since $f(x)$ is a polynomial, it is always continuous and differentiable. Focus on the $f(a)=f(b)$ condition." solution="Step 1: Check conditions for Rolle's Theorem.
The function $f(x) = (x-1)(x-2)(x-3)$ is a polynomial.

It is continuous on any closed interval $[a,b]$.

It is differentiable on any open interval $(a,b)$.

Therefore, we only need to check the third condition: $f(a) = f(b)$.

Step 2: Evaluate $f(x)$ at the endpoints of each given interval.

* For $[1, 2]$:
>

>
Since $f(1) = f(2) = 0$, Rolle's Theorem applies to $[1, 2]$.

* For $[2, 3]$:
>

>
Since $f(2) = f(3) = 0$, Rolle's Theorem applies to $[2, 3]$.

* For $[1, 3]$:
>

>
Since $f(1) = f(3) = 0$, Rolle's Theorem applies to $[1, 3]$.

* For $[0, 1]$:
>

>
Since $f(0) \neq f(1)$, Rolle's Theorem does not apply to $[0, 1]$.

Thus, Rolle's Theorem applies to the intervals $[1, 2]$, $[2, 3]$, and $[1, 3]$.
Answer: \boxed{[1, 2], [2, 3], [1, 3]}
"
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Summary

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What's Next?

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="For what value of $k$ is the function $f(x) = \begin{cases} \frac{x^2-4}{x-2} & x \ne 2 \\ k & x=2 \end{cases}$ continuous at $x=2$?" options=["1","2","3","4"] answer="4" hint="For continuity at a point, the limit of the function as x approaches that point must be equal to the function's value at that point." solution="For $f(x)$ to be continuous at $x=2$, we must have $\lim_{x \to 2} f(x) = f(2)$.
Given $f(2) = k$.
Now, we evaluate the limit:

Since $x \ne 2$, we can cancel $(x-2)$:

For continuity, $k$ must be equal to the limit. Thus, $k=4$.
Answer: \boxed{4}
"
:::

:::question type="MCQ" question="Let $f(x) = x|x|$. Which of the following statements about $f(x)$ is true at $x=0$?" options=["$f(x)$ is continuous but not differentiable." "$f(x)$ is differentiable but not continuous." "$f(x)$ is neither continuous nor differentiable." "$f(x)$ is both continuous and differentiable."] answer="$f(x)$ is both continuous and differentiable." hint="Evaluate the limit definition of the derivative at $x=0$. Remember that $f(x)$ can be written as a piecewise function." solution="The function $f(x) = x|x|$ can be written as:

Continuity at $x=0$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0$
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0$
$f(0) = 0^2 = 0$
Since $\lim_{x \to 0} f(x) = f(0) = 0$, $f(x)$ is continuous at $x=0$.

Differentiability at $x=0$:
We use the definition of the derivative: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.

As $h \to 0$, $|h| \to 0$.
So, $f'(0) = 0$.
Since the limit exists, $f(x)$ is differentiable at $x=0$.
Therefore, $f(x)$ is both continuous and differentiable at $x=0$.
Answer: \boxed{f(x) \text{ is both continuous and differentiable.}}
"
:::

:::question type="NAT" question="If $f(x) = x^2 - 2x$ on the interval $[0, 2]$, find the value of $c \in (0, 2)$ for which $f'(c) = 0$. (Apply Rolle's Theorem)" answer="1" hint="Rolle's Theorem states that if a function is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c)=0$. First, verify the conditions, then find $f'(x)$ and set it to zero." solution="First, verify the conditions for Rolle's Theorem:

$f(x) = x^2 - 2x$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$.

Check $f(a)$ and $f(b)$:

$f(0) = 0^2 - 2(0) = 0$
$f(2) = 2^2 - 2(2) = 4 - 4 = 0$
Since $f(0) = f(2)$, Rolle's Theorem applies.

Now, find $f'(x)$ and set it to zero:
$f'(x) = \frac{d}{dx}(x^2 - 2x) = 2x - 2$
Set $f'(c) = 0$:
$2c - 2 = 0 \implies 2c = 2 \implies c = 1$.
The value $c=1$ lies in the interval $(0, 2)$.
Answer: \boxed{1}
"
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What's Next?