Limits, Continuity, and Differentiability

Overview

In this chapter, we shall embark on a systematic study of the foundational principles of differential calculus: limits, continuity, and differentiability. These three concepts form the bedrock upon which the analysis of function behavior is built. We will see that they are not independent topics but rather form a logical hierarchy. The notion of a limit is prerequisite to a formal understanding of continuity, which in turn is a necessary condition for differentiability. A thorough command of this sequence is essential for modeling and solving a vast array of problems in engineering and the applied sciences.

For the Graduate Aptitude Test in Engineering (GATE), a mastery of this material is indispensable. The concepts explored herein are not merely theoretical exercises; they are fundamental to understanding more advanced topics in engineering mathematics, such as series expansions, numerical methods, and optimization algorithms which are central to computer science. Examination questions frequently test not only the mechanical application of formulae but also a nuanced conceptual grasp of the interplay between these three properties of functions. Our objective is, therefore, to build both computational proficiency and a deep, intuitive understanding of the subject matter.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Limits | Behavior of functions near a point |
| 2 | Continuity | Analyzing functions for uninterrupted graphs |
| 3 | Differentiability | Determining the instantaneous rate of change |

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Learning Objectives

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We now turn our attention to Limits...
## Part 1: Limits

Introduction

The concept of a limit is a foundational pillar of calculus, upon which the principles of continuity and differentiability are built. In engineering mathematics, a firm grasp of limits is not merely an academic exercise; it is essential for understanding the behavior of functions, analyzing rates of change, and approximating complex systems. A limit describes the value that a function "approaches" as its input approaches some specific value. It is concerned with the behavior of the function in the immediate vicinity of a point, not necessarily at the point itself.

We shall explore the formal definition of a limit, the properties that govern its algebraic manipulation, and, most critically for the GATE examination, the systematic techniques for evaluating limits. Our focus will be on resolving indeterminate forms—expressions such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$—which frequently appear in competitive examinations and require specific methodologies to determine the true limiting value. Mastery of these techniques is indispensable for solving problems in calculus efficiently and accurately.

x
f(x)

a
L
y = f(x)
(hole at x=a)

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Key Concepts

#
## 1. Indeterminate Forms

The simplest method for evaluating a limit is direct substitution. However, in many cases, substituting $x=a$ into $f(x)$ yields an expression that is mathematically undefined. Such expressions are known as indeterminate forms. Recognizing these forms is the first step toward selecting the correct evaluation technique.

The principal indeterminate forms are:

An indeterminate form does not imply that the limit does not exist. It simply means that the expression, in its current form, does not provide enough information to determine the limit. Algebraic manipulation or other advanced techniques are required to resolve the ambiguity.

#
## 2. Evaluation Technique: Rationalization

When a limit problem involves a radical (typically a square root) and results in the indeterminate form $\frac{0}{0}$, the method of rationalization is often effective. This technique involves multiplying the numerator and the denominator by the conjugate of the expression containing the radical. The conjugate of $(a+b)$ is $(a-b)$, and their product, $(a+b)(a-b) = a^2 - b^2$, eliminates the radical.

Worked Example:

Problem: Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$.

Solution:

Step 1: Attempt direct substitution.
Substituting $x=0$ gives $\frac{\sqrt{0+4}-2}{0} = \frac{2-2}{0}$, which results in the indeterminate form $\frac{0}{0}$.

Step 2: Identify the radical expression and its conjugate.
The expression is $(\sqrt{x+4} - 2)$. Its conjugate is $(\sqrt{x+4} + 2)$.

Step 3: Multiply the numerator and denominator by the conjugate.

Step 4: Simplify the numerator using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$.

Step 5: Cancel the common factor $x$ from the numerator and denominator.

Step 6: Evaluate the resulting limit by direct substitution.

Answer: $\frac{1}{4}$

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#
## 3. Evaluation Technique: L'Hôpital's Rule

L'Hôpital's Rule provides a powerful and often more direct method for evaluating limits that result in the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It states that under certain conditions, the limit of a quotient of two functions is equal to the limit of the quotient of their derivatives.

Worked Example (Revisiting the previous problem):

Problem: Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$ using L'Hôpital's Rule.

Solution:

Step 1: Verify the indeterminate form.
As established before, direct substitution yields $\frac{0}{0}$. Thus, L'Hôpital's Rule is applicable.

Step 2: Differentiate the numerator and the denominator separately.
Let $f(x) = \sqrt{x+4} - 2 = (x+4)^{1/2} - 2$. Then $f'(x) = \frac{1}{2}(x+4)^{-1/2} = \frac{1}{2\sqrt{x+4}}$.
Let $g(x) = x$. Then $g'(x) = 1$.

Step 3: Apply L'Hôpital's Rule.

Step 4: Evaluate the new limit by direct substitution.

Answer: $\frac{1}{4}$

We observe that both methods yield the same result, but L'Hôpital's Rule can be computationally faster for more complex functions.

#
## 4. Standard Limits

Certain limits appear so frequently that their results are considered standard. Memorizing these can significantly reduce calculation time in an exam.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="The value of the limit $\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ is ______." answer="3" hint="This is a 0/0 indeterminate form. You can either use factorization or L'Hôpital's Rule." solution="
Method 1: Factorization

Step 1: Substitute $x=2$ to check the form.

This is an indeterminate form.

Step 2: Factor the numerator and denominator.
The numerator is a difference of cubes: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$.
The denominator is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

Step 3: Cancel the common factor $(x-2)$.

Step 4: Evaluate by direct substitution.

Method 2: L'Hôpital's Rule

Step 1: The form is $\frac{0}{0}$, so the rule is applicable.

Step 2: Differentiate the numerator and denominator separately.
Derivative of $x^3-8$ is $3x^2$.
Derivative of $x^2-4$ is $2x$.

Step 3: Apply the rule.

Step 4: Evaluate by direct substitution.

Result: The value of the limit is 3.
"
:::

:::question type="MCQ" question="What is the value of the limit $\displaystyle \lim_{x \to 0} \frac{e^{4x} - 1}{\sin(2x)}$?" options=["0","1","2","4"] answer="2" hint="This is a 0/0 indeterminate form. Try to manipulate the expression to use standard limits or apply L'Hôpital's Rule." solution="
Method 1: Using Standard Limits

Step 1: The limit is of the form $\frac{e^0-1}{\sin(0)} = \frac{0}{0}$.

Step 2: We know the standard limits $\lim_{u \to 0} \frac{e^u-1}{u} = 1$ and $\lim_{v \to 0} \frac{\sin v}{v} = 1$. We can rewrite the expression to use these.

Step 3: Multiply and divide by $4x$ and $2x$.

Step 4: Separate the limits. As $x \to 0$, $4x \to 0$ and $2x \to 0$.

Step 5: Apply the standard limits.

Method 2: L'Hôpital's Rule

Step 1: The form is $\frac{0}{0}$.

Step 2: Differentiate numerator and denominator.
Derivative of $e^{4x}-1$ is $4e^{4x}$.
Derivative of $\sin(2x)$ is $2\cos(2x)$.

Step 3: Apply the rule.

Step 4: Evaluate by direct substitution.

Result: The correct option is 2.
"
:::

:::question type="NAT" question="Consider the limit $\displaystyle \lim_{x \to 5} \frac{x - 5}{\sqrt{3x+1} - 4}$. The value of this limit is (rounded to two decimal places) ______." answer="2.67" hint="Direct substitution leads to a 0/0 form. This problem involves a radical in the denominator, making rationalization an excellent choice. Alternatively, L'Hôpital's rule can be used." solution="
Method 1: Rationalization

Step 1: Substitute $x=5$ to find the form.

Step 2: Multiply the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{3x+1} + 4)$.

Step 3: Simplify the denominator.

Step 4: Factor the denominator and cancel the common term.

Step 5: Evaluate by direct substitution.

Step 6: Convert to a decimal rounded to two places.

Result: The value of the limit is 2.67.
"
:::

:::question type="MCQ" question="Evaluate $\displaystyle \lim_{x \to \infty} \left(\frac{x+3}{x-1}\right)^x$." options=["$e^2$","$e^4$","$e^{-2}$","$1$"] answer="$e^4$" hint="This limit is of the indeterminate form $1^\infty$. Rewrite the base as $(1 + f(x))$ and use the standard limit form $\lim_{u \to \infty} (1 + 1/u)^u = e$." solution="
Step 1: Identify the indeterminate form.
As $x \to \infty$, the base $\frac{x+3}{x-1} \to 1$ and the exponent $x \to \infty$. This is the $1^\infty$ form.

Step 2: Rewrite the base in the form $(1+f(x))$.

Step 3: Substitute this back into the limit expression.

Step 4: Manipulate the expression to match the standard form $\lim_{u \to \infty} (1 + k/u)^u = e^k$. A more direct approach is to make the exponent match the denominator of the fraction.
Let us set the exponent to be $\frac{x-1}{4}$ and then adjust.

Step 5: The inner part of the limit matches the form of $e$. As $x \to \infty$, let $u = \frac{x-1}{4}$, so $u \to \infty$. The inner limit becomes $\lim_{u \to \infty} (1+1/u)^u = e$.
So, the limit becomes:

Step 6: Evaluate the limit in the exponent.

Step 7: Combine the results.
The final result is $e^4$.

Result: The correct option is $e^4$.
"
:::

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Summary

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What's Next?

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Part 2: Continuity

Introduction

In the study of calculus, the concept of a limit provides the foundational language for describing the behavior of a function as its input approaches a certain value. Building directly upon this foundation, we introduce the notion of continuity. Intuitively, a function is continuous if its graph can be drawn as a single, unbroken curve. This means there are no abrupt jumps, holes, or vertical asymptotes within its domain.

While this graphical interpretation is useful, a more rigorous mathematical definition is required for formal analysis. Continuity at a point ensures that the function's value at that point is precisely what we would expect it to be, based on its values at nearby points. This property is fundamental to many important theorems in calculus, including the Intermediate Value Theorem and the Extreme Value Theorem, and serves as a necessary prerequisite for the concept of differentiability, a topic of significant importance in the GATE examination.

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Key Concepts

#
## 1. The Three Conditions for Continuity

For a function to be continuous at a point $x=c$, all three conditions from the definition must be met. The failure of even one condition implies that the function is discontinuous at that point. Let us examine the second condition more closely. The existence of the limit $\lim_{x \to c} f(x)$ itself requires that the left-hand limit (LHL) and the right-hand limit (RHL) are equal.

For the limit to exist, we must have $LHL = RHL$. Therefore, the test for continuity can be expressed concisely.

Worked Example:

Problem: Determine if the function $f(x)$ defined below is continuous at $x = 2$.

Solution:

We must check the three conditions for continuity at $x = 2$.

Step 1: Evaluate the function at the point $x=2$.

From the function definition, when $x=2$, $f(x) = 3$.

The function is defined at $x=2$.

Step 2: Calculate the left-hand limit (LHL) as $x \to 2^-$.

For $x < 2$, the function is defined as $f(x) = x^2 - 1$.

Step 3: Calculate the right-hand limit (RHL) as $x \to 2^+$.

For $x > 2$, the function is defined as $f(x) = x + 1$.

Step 4: Compare the LHL, RHL, and the function value.

We have found that $LHL = 3$, $RHL = 3$, and $f(2) = 3$.

Answer: Since all three values are equal, the function $f(x)$ is continuous at $x = 2$.

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#
## 2. Types of Discontinuities

When a function fails to be continuous at a point, it is said to have a discontinuity at that point. Understanding the nature of these discontinuities is important.

x
y
Continuous Function

x
y
Jump Discontinuity

c

Removable Discontinuity: This occurs at $x=c$ if $\lim_{x \to c} f(x)$ exists, but is not equal to $f(c)$, or if $f(c)$ is not defined at all. It is called "removable" because the discontinuity can be eliminated by redefining the function at the single point $x=c$. This corresponds to a "hole" in the graph.

Jump Discontinuity: This occurs at $x=c$ if the left-hand and right-hand limits both exist as finite numbers, but are not equal to each other ($\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$). The graph "jumps" from one value to another.

Infinite Discontinuity: This occurs at $x=c$ if at least one of the one-sided limits is infinite (i.e., $\lim_{x \to c^-} f(x) = \pm\infty$ or $\lim_{x \to c^+} f(x) = \pm\infty$). This typically corresponds to a vertical asymptote in the graph.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="For the function $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases}$ to be continuous at $x=3$, what must be the value of $k$?" answer="6" hint="For continuity, the value of the function at the point must equal the limit at that point. Simplify the expression for $x \neq 3$ first." solution="
Step 1: For the function to be continuous at $x=3$, we must have $\lim_{x \to 3} f(x) = f(3)$.

Step 2: From the definition, we know that $f(3) = k$.

Step 3: Now, we compute the limit. For $x \neq 3$, the function is $f(x) = \frac{x^2 - 9}{x - 3}$.

Step 4: This is an indeterminate form $\frac{0}{0}$. We can simplify the expression by factoring the numerator.

Step 5: Since $x \to 3$ implies $x \neq 3$, we can cancel the $(x-3)$ term.

Step 6: Equate the limit with the function value.

Result: The value of $k$ must be 6.
"
:::

:::question type="MCQ" question="The function $f(x) = \frac{|x|}{x}$ has which type of discontinuity at $x=0$?" options=["Removable Discontinuity","Jump Discontinuity","Infinite Discontinuity","The function is continuous at x=0"] answer="Jump Discontinuity" hint="Evaluate the left-hand limit and the right-hand limit at $x=0$ separately. Recall the definition of $|x|$." solution="
Step 1: Analyze the function $f(x) = \frac{|x|}{x}$. We need to check for continuity at $x=0$.

Step 2: First, check if $f(0)$ is defined. The function is not defined at $x=0$ because it results in division by zero. This immediately tells us the function is discontinuous at $x=0$. Now we must classify the discontinuity.

Step 3: Calculate the left-hand limit (LHL) as $x \to 0^-$. For $x < 0$, $|x| = -x$.

Step 4: Calculate the right-hand limit (RHL) as $x \to 0^+$. For $x > 0$, $|x| = x$.

Step 5: Compare the LHL and RHL.

Since both limits are finite but not equal ($LHL \neq RHL$), the function has a jump discontinuity at $x=0$.

Result: The correct option is Jump Discontinuity.
"
:::

:::question type="MSQ" question="Which of the following functions are continuous for all $x \in \mathbb{R}$?" options=["$f(x) = \tan(x)$","$f(x) = e^{-x^2}$","$f(x) = \lfloor x \rfloor$ (floor function)","$f(x) = x^3 - 2x + 1$"] answer="f(x) = e^{-x^2},f(x) = x^3 - 2x + 1" hint="Consider the domain of each function and points where they might have breaks, jumps, or asymptotes. Polynomials and compositions of continuous functions are continuous." solution="

• $f(x) = \tan(x)$: The tangent function is defined as $\frac{\sin(x)}{\cos(x)}$. It has infinite discontinuities wherever $\cos(x) = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. Therefore, it is not continuous for all $x \in \mathbb{R}$.

• $f(x) = e^{-x^2}$: This is a composition of two continuous functions. The function $g(x) = -x^2$ is a polynomial and is continuous everywhere. The function $h(u) = e^u$ is an exponential function and is continuous everywhere. The composition $h(g(x)) = e^{-x^2}$ is therefore continuous for all $x \in \mathbb{R}$.

• $f(x) = \lfloor x \rfloor$: The floor function (or greatest integer function) gives the greatest integer less than or equal to $x$. This function has jump discontinuities at every integer value. For example, at $x=2$, $\lim_{x \to 2^-} \lfloor x \rfloor = 1$ while $\lfloor 2 \rfloor = 2$. It is not continuous for all $x \in \mathbb{R}$.

• $f(x) = x^3 - 2x + 1$: This is a polynomial function. All polynomial functions are continuous for all real numbers.

Result: The functions $f(x) = e^{-x^2}$ and $f(x) = x^3 - 2x + 1$ are continuous for all $x \in \mathbb{R}$.
"
:::

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Summary

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What's Next?

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Part 3: Differentiability

Introduction

In the study of calculus, the concept of a derivative stands as a cornerstone, providing a powerful tool for analyzing the behavior of functions. Differentiability, the property of a function possessing a derivative at a point, is of profound importance. Geometrically, it signifies the existence of a unique, non-vertical tangent line to the function's graph at that point, indicating a "smoothness" in the curve. Analytically, it quantifies the instantaneous rate of change of the function with respect to its variable.

For the GATE examination, a firm grasp of differentiability is essential. It is not merely a procedural skill of applying differentiation formulas, but a deeper understanding of the conditions under which a derivative exists. This includes the intricate relationship between differentiability and continuity, and the methods for analyzing functions that are defined piecewise or through operations such as `max` and `min`. We shall explore these foundational concepts, providing the necessary theoretical framework and problem-solving techniques for success.

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Key Concepts

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## 1. Left-Hand and Right-Hand Derivatives

For the limit in the definition of the derivative to exist, the limit must be the same whether $h$ approaches $0$ from the positive side or the negative side. This gives rise to the concepts of Left-Hand and Right-Hand Derivatives.

The Right-Hand Derivative (RHD) at $x=c$ is defined as:

The Left-Hand Derivative (LHD) at $x=c$ is defined as:

A function $f(x)$ is differentiable at $x=c$ if and only if both the LHD and RHD exist, are finite, and are equal to each other.

Worked Example:

Problem: Investigate the differentiability of the function $f(x) = |x-2|$ at the point $x=2$.

Solution:

We must compute the Left-Hand and Right-Hand Derivatives at $x=2$.

Step 1: Calculate the Right-Hand Derivative (RHD).
For $x > 2$, $f(x) = x-2$. We use the definition of the RHD at $c=2$.

Since $h \to 0^+$, $h$ is positive, so $|h| = h$.

Step 2: Calculate the Left-Hand Derivative (LHD).
For $x < 2$, $f(x) = -(x-2) = 2-x$. We use the definition of the LHD at $c=2$.

Since $h \to 0^-$, $h$ is negative, so $|h| = -h$.

Step 3: Compare LHD and RHD.

We observe that $L f'(2) = -1$ and $R f'(2) = 1$.

Answer: Since the Left-Hand and Right-Hand Derivatives at $x=2$ are not equal, the function $f(x) = |x-2|$ is not differentiable at $x=2$.

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#
## 2. Differentiability and Continuity

A fundamental theorem connects the concepts of differentiability and continuity. It provides a necessary condition for a function to be differentiable.

This principle is visually intuitive. A differentiable function is "smooth" at a point, which implies it must be connected (continuous). However, a function can be connected but have a sharp corner or a cusp, at which point it is not smooth and thus not differentiable.

x
y


Differentiable
(Smooth)




x
y



Continuous, Not Differentiable
(Sharp Corner)

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#
## 3. Differentiability of Piecewise Functions

Piecewise functions are a common subject in GATE questions on differentiability. The analysis must be performed at the boundary points where the function definition changes.

To check for differentiability of a piecewise function at a boundary point $x=c$:

Check for Continuity: First, ensure the function is continuous at $x=c$. This requires that the limit from the left, the limit from the right, and the function value are all equal.

If the function is not continuous at $c$, it cannot be differentiable at $c$.

Check for Differentiability: If the function is continuous, proceed to check if the derivatives are equal. Calculate the LHD and RHD at $x=c$ by differentiating the respective pieces of the function and evaluating them at $c$.

Worked Example:

Problem:
Consider the function

Find the values of $a$ and $b$ such that the function is differentiable for all $x \in \mathbb{R}$.

Solution:

The function is composed of polynomials, which are differentiable everywhere except possibly at the boundary point $x=1$. We must enforce continuity and differentiability at $x=1$.

Step 1: Enforce the continuity condition at $x=1$.
The left-hand limit must equal the right-hand limit.

Step 2: Enforce the differentiability condition at $x=1$.
The left-hand derivative must equal the right-hand derivative. First, we find the derivatives of the pieces.

Now, we equate the LHD and RHD at $x=1$.

Step 3: Solve for $b$ using the result from Step 1.
Substitute $a=4$ into equation (1).

Answer: For the function to be differentiable everywhere, we must have $a = 4$ and $b = -1$.

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#
## 4. Differentiability of `max`, `min`, and Absolute Value Functions

Functions defined using `max`, `min`, or absolute value often have points of non-differentiability. These points typically occur where the arguments of the `max` or `min` function are equal, as this is where the function's definition can switch, potentially creating a "corner".

Consider a function $f(x) = \max\{g(x), h(x)\}$. The points of interest are the solutions to the equation $g(x) = h(x)$. Let $c$ be such a point. To check for differentiability at $c$, we must evaluate the LHD and RHD. In a neighborhood to the left of $c$, $f(x)$ will be either $g(x)$ or $h(x)$, and to the right, it might switch. The LHD will be either $g'(c)$ or $h'(c)$, and similarly for the RHD. The function is differentiable at $c$ only if $g(c)=h(c)$ and $g'(c)=h'(c)$.

Worked Example:

Problem: Find the points where the function $f(x) = \min\{x^2, 2-x\}$ is not differentiable.

Solution:

Step 1: Find the points of intersection of the two functions.
The points where the function definition might change are where $x^2 = 2-x$.

The potential points of non-differentiability are $x=-2$ and $x=1$.

Step 2: Analyze the function definition around the intersection points.

• For $x < -2$, we can test a point like $x=-3$. $g(-3) = (-3)^2 = 9$ and $h(-3) = 2 - (-3) = 5$. Since $h(x) < g(x)$, $f(x) = h(x) = 2-x$.


• For $-2 < x < 1$, we can test $x=0$. $g(0) = 0^2 = 0$ and $h(0) = 2-0=2$. Since $g(x) < h(x)$, $f(x) = g(x) = x^2$.


• For $x > 1$, we can test $x=2$. $g(2) = 2^2 = 4$ and $h(2) = 2-2=0$. Since $h(x) < g(x)$, $f(x) = h(x) = 2-x$.

So, the piecewise definition of $f(x)$ is:

Step 3: Check differentiability at $x = -2$.
The derivatives of the pieces are $g'(x) = 2x$ and $h'(x) = -1$.

Since $L f'(-2) \neq R f'(-2)$, the function is not differentiable at $x=-2$.

Step 4: Check differentiability at $x = 1$.

Since $L f'(1) \neq R f'(1)$, the function is not differentiable at $x=1$.

Answer: The function is not differentiable at the points $x=-2$ and $x=1$.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="Let the function $f(x)$ be defined as

If $f(x)$ is differentiable at $x=0$, then the value of $a$ is _______." answer="1" hint="For differentiability at $x=0$, the function must first be continuous at $x=0$. Use this to relate $a$ and $b$. Then, equate the left-hand and right-hand derivatives at $x=0$." solution="
Step 1: Enforce continuity at $x=0$.

Step 2: Enforce differentiability at $x=0$.
First, find the derivatives of the pieces.
For $x < 0$, $f'(x) = e^x$.
For $x > 0$, $f'(x) = -a\sin(x)$.

Now, equate the LHD and RHD at $x=0$.

This is a contradiction. Let's re-read the problem. Ah, I see a common trap in problem creation. Let me adjust the problem slightly to make it solvable. Let's make the function $e^x + ax$.

Corrected Question for a valid solution:
Let the function $f(x)$ be defined as

If $f(x)$ is differentiable at $x=0$, then the value of $a$ is _______.

Solution to Corrected Question:

Step 1: Enforce continuity at $x=0$.

Step 2: Enforce differentiability at $x=0$.
The derivatives are:
For $x < 0$, $f'(x) = 2$.
For $x > 0$, $f'(x) = a\cos(x)$.
Equating LHD and RHD at $x=0$:

Final Answer for Corrected Question: 2.
Let me create a new, clean question for the user based on the original structure.

Actual Practice Question:
:::question type="NAT" question="Let the function $f(x)$ be defined as

If $f(x)$ is differentiable at $x=1$, then the value of $a$ is _______." answer="0.5" hint="First, apply the continuity condition at $x=1$. Then, apply the differentiability condition by equating the left-hand and right-hand derivatives at $x=1$. This will give you a simple equation to solve for $a$." solution="
Step 1: Apply the continuity condition at $x=1$.
The function value from the left must equal the function value from the right.




This equation is $1=1$, which is always true. This means the function is continuous for any value of $a$. We must use the differentiability condition to find $a$.

Step 2: Apply the differentiability condition at $x=1$.
First, find the derivatives of the two pieces.
For $x < 1$, $f'(x) = \frac{d}{dx}(x+a) = 1$.
For $x > 1$, $f'(x) = \frac{d}{dx}(ax^2+1) = 2ax$.

Now, equate the LHD and RHD at $x=1$.

Result: The value of $a$ is $0.5$.
"
:::

:::question type="MCQ" question="The set of all points where the function $f(x) = \max\{x^2, -x^2+2\}$ is NOT differentiable is" options=["$\\{1\\}$", "$\\{-1, 1\\}$", "$\\{0\\}$", "$\\{-1, 0, 1\\}$"] answer="$\\{-1, 1\\}$" hint="Find the points where the two functions are equal: $x^2 = -x^2+2$. Then, check if the derivatives of the two functions are equal at these points." solution="
Step 1: Find the points of intersection.
Set the arguments of the $\max$ function equal to each other.

These are the potential points of non-differentiability.

Step 2: Find the derivatives of the individual functions.
Let $g(x) = x^2$ and $h(x) = -x^2+2$.

Step 3: Check differentiability at $x=1$.
We compare the derivatives at this point.

Since $g'(1) \neq h'(1)$, the function is not differentiable at $x=1$.

Step 4: Check differentiability at $x=-1$.
We compare the derivatives at this point.

Since $g'(-1) \neq h'(-1)$, the function is not differentiable at $x=-1$.

Result: The function is not differentiable at the points $\\{-1, 1\\}$.
"
:::

:::question type="MSQ" question="Consider the function $f(x) = |x^2 - 4|$. Which of the following statements is/are TRUE?" options=["$f(x)$ is continuous at $x=2$.", "$f(x)$ is differentiable at $x=2$.", "$f(x)$ is differentiable at $x=0$.", "The left-hand derivative of $f(x)$ at $x=2$ is $-4$."] answer="A,C,D" hint="Rewrite the function in piecewise form. The critical points are where $x^2-4=0$. Check continuity and differentiability at these points, as well as at other points like $x=0$." solution="
The function is $f(x) = |x^2 - 4|$. The expression inside the absolute value is zero when $x^2-4=0$, i.e., at $x=2$ and $x=-2$.
The piecewise definition of the function is:

Option A: $f(x)$ is continuous at $x=2$.
The function is a composition of a polynomial and the absolute value function, both of which are continuous everywhere. Therefore, $f(x)$ is continuous everywhere, including at $x=2$. This statement is TRUE.

Option B: $f(x)$ is differentiable at $x=2$.
We must check if LHD = RHD at $x=2$.
The derivative of the right piece ($x^2-4$) is $2x$. So, $R f'(2) = 2(2) = 4$.
The derivative of the left piece ($4-x^2$) is $-2x$. So, $L f'(2) = -2(2) = -4$.
Since $L f'(2) \neq R f'(2)$, the function is not differentiable at $x=2$. This statement is FALSE.

Option C: $f(x)$ is differentiable at $x=0$.
At $x=0$, the function is defined as $f(x) = 4-x^2$. This is a simple polynomial, which is differentiable everywhere in its domain. The derivative is $f'(x)=-2x$, and $f'(0)=0$. The function is differentiable at $x=0$. This statement is TRUE.

Option D: The left-hand derivative of $f(x)$ at $x=2$ is $-4$.
As calculated for Option B, the LHD at $x=2$ corresponds to the derivative of $4-x^2$ evaluated at $x=2$, which is $-2x \rvert_{x=2} = -4$. This statement is TRUE.

Result: The correct options are A, C, and D.
"
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Summary

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What's Next?

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Consider the function $f(x) = \begin{cases} ax^2 + b & \text{if } x \leq 1 \\ 2x - 1 & \text{if } x > 1 \end{cases}$. For what values of $a$ and $b$ is the function $f(x)$ differentiable at $x=1$?" options=["$a=1, b=0$","$a=1, b=1$","$a=2, b=-1$","$a=0, b=1$"] answer="A" hint="For a function to be differentiable at a point, it must first be continuous at that point. Set the Left-Hand Limit equal to the Right-Hand Limit for continuity, and the Left-Hand Derivative equal to the Right-Hand Derivative for differentiability." solution="For $f(x)$ to be differentiable at $x=1$, it must first be continuous at $x=1$.

Condition for Continuity:
The Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL) at $x=1$.

Condition for Differentiability:
The Left-Hand Derivative (LHD) must equal the Right-Hand Derivative (RHD) at $x=1$.
First, let's find the derivatives of the pieces of the function:

Now, we evaluate the LHD and RHD at $x=1$:


Equating the two:

Substituting $a=1$ into equation (1):

Thus, the function is differentiable at $x=1$ for $a=1$ and $b=0$.
"
:::

:::question type="NAT" question="The value of the limit $\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{1/x^2}$ is:" answer="0.569" hint="This is an indeterminate form of the type $1^\infty$. Let $L$ be the limit and take the natural logarithm of both sides. Then use L'Hôpital's rule on the resulting expression." solution="Let $L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{1/x^2}$.
This is of the indeterminate form $1^\infty$. To solve this, we take the natural logarithm of both sides:

Since the natural logarithm is a continuous function, we can move the limit outside:


This is of the form $\frac{0}{0}$. We can use the Taylor series expansion for $\sin x$ and $\ln(1-u)$.
The Taylor series for $\sin x$ around $x=0$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$

Now, we substitute this into our expression for $\ln L$:

Using the Taylor expansion for $\ln(1-u) \approx -u$ for small $u$, we let $u = \frac{x^2}{6} - O(x^4)$.



Since $\ln L = -1/6$, we have $L = e^{-1/6}$.

Wait, let's re-check the calculation using L'Hôpital's rule directly as it's more robust.

This is in the $\frac{0}{0}$ form. Applying L'Hôpital's Rule:

This is still $\frac{0}{0}$. Let's simplify the denominator using $\sin x \approx x$ for $x \to 0$:

Apply L'Hôpital's Rule again:

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

So, $L = e^{-1/6}$.
Calculating the value: $e^{-1/6} \approx 0.84648$.
Let's re-read the question. The answer format is NAT. Let's provide the answer rounded to 3 decimal places.
$0.846$.

There seems to be a discrepancy in my initial thought vs the solution. Let me re-verify.
$L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{1/x^2}$.
$\ln L = \lim_{x \to 0} \frac{\ln(\frac{\sin x}{x})}{x^2}$.
Let's use the Taylor series again, it's often faster.
$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots = 1 - \frac{x^2}{6} + O(x^4)$.
$\ln(1+u) = u - \frac{u^2}{2} + \dots$.
Let $u = -\frac{x^2}{6} + O(x^4)$.
$\ln(\frac{\sin x}{x}) = \ln(1 - \frac{x^2}{6} + O(x^4)) = (-\frac{x^2}{6} + O(x^4)) - \frac{1}{2}(-\frac{x^2}{6} + O(x^4))^2 + \dots$
$= -\frac{x^2}{6} + O(x^4)$.
So, $\ln L = \lim_{x \to 0} \frac{-\frac{x^2}{6} + O(x^4)}{x^2} = \lim_{x \to 0} (-\frac{1}{6} + O(x^2)) = -\frac{1}{6}$.
$L = e^{-1/6} \approx 0.84648$. The answer should be 0.846.

Let me try a different problem to avoid any ambiguity.
How about $\lim_{x \to 0} \frac{x - \sin(x)}{1 - \cos(x)}$.
This is $\frac{0}{0}$.
L'Hôpital: $\lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x)}$. Still $\frac{0}{0}$.
L'Hôpital again: $\lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \frac{0}{1} = 0$. This is too simple.

Let's stick with the original problem but make sure the answer is correct. $e^{-1/6}$ seems correct. Let me use a calculator to be sure. $e^{-1/6} \approx 0.8464817...$. Rounded to three decimal places, it's 0.846.
The question is good, so I'll keep it. The NAT answer should be 0.846.

Final check of the steps.

Identify form $1^\infty$.

Take log: $\ln L = \lim \frac{1}{x^2} \ln(\frac{\sin x}{x})$.

This is form $\frac{0}{0}$.

Apply L'Hôpital's.

Numerator derivative: $\frac{d}{dx} \ln(\frac{\sin x}{x}) = \frac{x}{\sin x} \cdot \frac{x \cos x - \sin x}{x^2} = \frac{x \cos x - \sin x}{x \sin x}$.
Denominator derivative: $2x$.
Limit becomes: $\lim_{x \to 0} \frac{x \cos x - \sin x}{2x^2 \sin x}$.

Using Taylor series $\sin x \approx x$ in the denominator makes it $\lim_{x \to 0} \frac{x \cos x - \sin x}{2x^3}$.

Apply L'Hôpital's again.

Numerator derivative: $1 \cdot \cos x + x(-\sin x) - \cos x = -x \sin x$.
Denominator derivative: $6x^2$.
Limit becomes: $\lim_{x \to 0} \frac{-x \sin x}{6x^2} = \lim_{x \to 0} \frac{-\sin x}{6x} = -\frac{1}{6}$.

So $\ln L = -1/6$, which means $L = e^{-1/6}$.

The solution is correct. The answer is approximately 0.846. I will use this.
Wait, the prompt says "NAT answers must be plain numbers". Is 0.846 a plain number? Yes. Is it possible GATE asks for an exact form sometimes? No, NAT is for numerical values. Okay, 0.846 is fine.

Let's create two more questions. One MSQ and one more MCQ.
MSQ Idea:
Give a function, maybe $f(x) = x|x|$ or $f(x) = |x-1| + |x+1|$.
Let's use $f(x) = |x^2 - 4|$ on the interval $[-3, 3]$.
Ask which statements are true.
A) $f(x)$ is continuous on $[-3, 3]$. (True, absolute value of a polynomial is continuous).
B) $f(x)$ is differentiable on $(-3, 3)$. (False, not differentiable at $x=2, x=-2$).
C) Rolle's Theorem is applicable to $f(x)$ on $[-3, 3]$. (False, not differentiable on the open interval. Also check $f(-3)$ and $f(3)$. $f(-3) = |9-4|=5$. $f(3)=|9-4|=5$. So $f(a)=f(b)$ holds, but differentiability fails).
D) Lagrange's Mean Value Theorem guarantees a point $c \in (-3, 3)$ such that $f'(c) = 0$. (False, LMVT requires differentiability on the open interval).
This is a good MSQ. It tests continuity, differentiability, and the conditions for MVT.
The correct options would be just A. Let me re-think. An MSQ should ideally have more than one correct answer.
Let's change the function or the options.
Let $f(x) = x^3 - 3x^2 + 2x$ on $[0, 2]$.
A) $f(x)$ is continuous on $[0, 2]$. (True, it's a polynomial).
B) $f(x)$ is differentiable on $(0, 2)$. (True, it's a polynomial).
C) Rolle's Theorem is applicable on $[0, 2]$. (Check conditions: continuous on [0,2], differentiable on (0,2). $f(0) = 0$. $f(2) = 8 - 12 + 4 = 0$. So $f(0)=f(2)$. All conditions met. True).
D) There exists a $c \in (0, 2)$ such that $f'(c) = 1$. (LMVT guarantees $f'(c) = \frac{f(2)-f(0)}{2-0} = \frac{0-0}{2}=0$. This is Rolle's. But does a point exist where the slope is 1? $f'(x) = 3x^2 - 6x + 2$. Set $f'(x)=1 \implies 3x^2 - 6x + 1 = 0$. $x = \frac{6 \pm \sqrt{36-12}}{6} = 1 \pm \frac{\sqrt{24}}{6} = 1 \pm \frac{2\sqrt{6}}{6} = 1 \pm \frac{\sqrt{6}}{3}$. $\sqrt{6} \approx 2.45$. So $x \approx 1 \pm 0.816$. Both $1.816$ and $0.184$ are in $(0,2)$. So this statement is also True).
This is a good MSQ. Correct answers: A, B, C, D. All are correct.

Final MCQ Idea:
A question on the type of discontinuity.
Let $f(x) = \frac{x^2 - 4}{x-2}$ for $x \neq 2$ and $f(2)=5$.
What kind of discontinuity does $f(x)$ have at $x=2$?
Options: A) Removable, B) Jump, C) Infinite, D) Oscillating.
Solution: $\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4$.
The limit exists and is finite (value is 4). However, $f(2)=5$. Since $\lim_{x \to 2} f(x) \neq f(2)$, it's a removable discontinuity. If we were to redefine $f(2)=4$, the function would become continuous.
This is a classic and good question.

So the four questions are:

MCQ: Differentiability of a piecewise function.

NAT: Limit of the form $1^\infty$.

MSQ: Properties of a polynomial function on an interval (Continuity, Differentiability, Rolle's, LMVT application).

MCQ: Type of discontinuity.

This set of questions provides good coverage of the chapter's topics.

What's Next Section:
I will structure this part as planned.

• Connection to Previous Learning: Basic functions, algebra.


• Builds Towards:

- Applications of Derivatives (Maxima/Minima, Rate of change)
- Integration (Fundamental Theorem of Calculus)
- Series Expansions (Taylor and Maclaurin Series)
- Differential Equations
- Vector Calculus (Partial Derivatives) - this is a good point to add.

The overall structure and content seem complete and well-aligned with the prompt. I will now write the final response. I will be careful with the LaTeX formatting and the specific markdown for the question blocks.
Final check on NAT answer. The prompt asks for a plain number. Should I use 0.846 or the exact $e^{-1/6}$? The hint suggests calculation and the type is "NAT" (Numerical Answer Type). Numerical answers are expected to be decimals. Rounding to 3 decimal places is standard practice unless specified otherwise. So 0.846 is the correct format. Let me change the answer to 0.846 to be safe.
Wait, my previous NAT problem had a potential issue. What if GATE expects an exact answer expressed as a number? $e^{-1/6}$ is not a number that can be typed in. A problem like $\lim_{x \to \infty} (1 + \frac{2}{x})^{3x}$ would be better.
$L = \lim_{x \to \infty} (1 + \frac{2}{x})^{3x}$. This is of the form $1^\infty$.
We know the standard form $\lim_{x \to \infty} (1 + \frac{a}{x})^{bx} = e^{ab}$.
Here, $a=2, b=3$. So the limit is $e^{2 \cdot 3} = e^6$.
This is also not a numerical answer.

Let's find a limit that evaluates to a clean integer or rational number.
$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.
Form is $\frac{0}{0}$.
L'Hôpital: $\lim_{x \to 0} \frac{e^x - 1}{2x}$.
Still $\frac{0}{0}$.
L'Hôpital again: $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2} = 0.5$.
This is a much better NAT question. The answer is a clean decimal. I will use this instead.

So the questions are now:

MCQ: Differentiability of piecewise function (good).

NAT: Limit using L'Hôpital's rule evaluating to 0.5 (better).

MSQ: Properties of a polynomial on an interval (good).

MCQ: Type of discontinuity (good).

This set is strong. I'm ready to write the final output.