Continuity

This chapter establishes the rigorous definition and implications of continuity, a foundational concept in calculus critical for understanding differentiability and integrability. Mastery of continuity at a point, across intervals, and for piecewise functions, alongside the Intermediate Value Theorem, is essential for advanced problem-solving and success in CMI examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Continuity at a point | | 2 | Interval continuity | | 3 | Continuity of piecewise functions | | 4 | Intermediate value reasoning |

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We begin with Continuity at a point.

Part 1: Continuity at a point

Continuity at a Point

Overview

Continuity at a point is one of the most important local ideas in calculus. A function is continuous at a point if there is no break, jump, hole, or mismatch there. In exam questions, continuity is often tested through piecewise functions, modulus functions, rational expressions, and parameter finding. The real skill is to compare three things carefully:

• the left-hand limit

• the right-hand limit

• the actual function value

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Learning Objectives

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Core Definition

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Three-Step Test

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What Can Go Wrong

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Common Cases

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Piecewise Functions

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Removable Discontinuity

Example idea: $\qquad f(x)=\dfrac{x^2-1}{x-1}$ for $x\ne 1$ For $x\ne 1$, this simplifies to $\qquad f(x)=x+1$ So $\qquad \lim_{x\to 1} f(x)=2$ But if $f(1)$ is not defined, then the function is not continuous at $x=1$. ---

Jump Discontinuity

This usually appears in piecewise functions. ---

Minimal Worked Examples

Example 1 Check continuity of $\qquad f(x)= \begin{cases} x+2, & x<1 \\ 4, & x=1 \\ 2x+1, & x>1 \end{cases} $ at $x=1$. Left-hand limit: $\qquad \lim_{x\to 1^-} f(x)=1+2=3$ Right-hand limit: $\qquad \lim_{x\to 1^+} f(x)=2(1)+1=3$ Function value: $\qquad f(1)=4$ Since the common limit is $3$ but $f(1)=4$, the function is not continuous at $x=1$. --- Example 2 Check continuity of $|x|$ at $x=0$. Left-hand limit: $\qquad \lim_{x\to 0^-}|x|=0$ Right-hand limit: $\qquad \lim_{x\to 0^+}|x|=0$ Function value: $\qquad |0|=0$ So $|x|$ is continuous at $x=0$. ---

Algebraic Simplification in Limit-Based Continuity

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Graph Interpretation

But remember: the graph idea helps intuition, while the algebraic limit definition gives the proof. ::: ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is sufficient for a function $f$ to be continuous at $x=a$?" options=["$f(a)$ is defined","The left-hand limit exists","$\lim_{x\to a}f(x)$ exists and equals $f(a)$","$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$"] answer="C" hint="Continuity requires the limit to match the function value." solution="A function $f$ is continuous at $x=a$ exactly when $\qquad \lim_{x\to a} f(x)=f(a)$ This automatically includes:

• existence of $f(a)$

• existence of the limit

• equality of the two

So the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Let $\qquad f(x)= \begin{cases} x^2+1, & x<2 \\ k, & x=2 \\ 3x-1, & x>2 \end{cases} $ Find the value of $k$ for which $f$ is continuous at $x=2$." answer="5" hint="Match the left-hand limit, right-hand limit, and function value." solution="For continuity at $x=2$, we need $\qquad \lim_{x\to 2^-} f(x)=f(2)=\lim_{x\to 2^+} f(x)$ Left-hand limit: $\qquad \lim_{x\to 2^-}(x^2+1)=2^2+1=5$ Right-hand limit: $\qquad \lim_{x\to 2^+}(3x-1)=3(2)-1=5$ So the common limit is $5$. Therefore we must have $\qquad k=5$ Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $\lim_{x\to a} f(x)$ exists and equals $f(a)$, then $f$ is continuous at $a$","If $f(a)$ is defined, then $f$ is continuous at $a$","If $\lim_{x\to a^-}f(x)\ne \lim_{x\to a^+}f(x)$, then $f$ is not continuous at $a$","A rational function is continuous at every point where its denominator is nonzero"] answer="A,C,D" hint="Use the definition and standard continuity facts." solution="1. True. This is exactly the definition of continuity at a point.

False. Merely being defined at a point does not guarantee continuity.

True. If the left and right limits differ, then the two-sided limit does not exist, so continuity fails.

True. Rational functions are continuous wherever the denominator is nonzero.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Examine the continuity at $x=1$ of the function $\qquad f(x)= \begin{cases} \dfrac{x^2-1}{x-1}, & x\ne 1 \\ 3, & x=1 \end{cases} $" answer="Continuous at$x=1$" hint="Simplify the rational expression for$x\ne 1$." solution="For$x\ne 1$, $\qquad \dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1$ So $\qquad \lim_{x\to 1} \dfrac{x^2-1}{x-1}=\lim_{x\to 1}(x+1)=2$ But the function value is given as $\qquad f(1)=3$ Since $\qquad \lim_{x\to 1} f(x)=2 \ne 3=f(1)$, the function is not continuous at $x=1$. So the correct conclusion is: $\qquad \boxed{\text{The function is not continuous at } x=1.}$" ::: ---

Summary

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Part 2: Interval continuity

Interval Continuity

Overview

Continuity on an interval means a function behaves without breaks, jumps, or missing points throughout that interval. In calculus, this topic is important because many major theorems, such as the Intermediate Value Theorem and Extreme Value Theorem, require continuity on suitable intervals. In CMI-style problems, interval continuity is often tested through piecewise functions, endpoint behaviour, and careful treatment of open and closed intervals. ---

Learning Objectives

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Continuity at a Point

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Continuity on an Open Interval

This means that for every interior point, the usual two-sided limit condition must hold. ---

Continuity on a Closed Interval

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Continuity on Half-Open Intervals

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One-Sided Continuity

These are essential for endpoint continuity and piecewise definitions. ---

Standard Continuous Functions

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Piecewise Functions and Interval Continuity

For a joining point $x=c$, compare:

• left-hand limit

• right-hand limit

• function value

If all agree, the function is continuous there. ---

Minimal Worked Examples

Example 1 Is the function $\qquad f(x)=\dfrac{x^2-1}{x-1}$ continuous on $(1,\infty)$? For $x\ne1$, $\qquad \dfrac{x^2-1}{x-1}=x+1$ Since $x=1$ is not in the interval $(1,\infty)$, the function is defined and continuous for every point of that interval. Hence it is continuous on $\boxed{(1,\infty)}$. --- Example 2 Check continuity of $\qquad f(x)= \begin{cases} x+1,& x<2 \\ 5,& x=2 \\ 3x-1,& x>2 \end{cases}$ at $x=2$. Left-hand limit: $\qquad \lim_{x\to2^-}f(x)=2+1=3$ Right-hand limit: $\qquad \lim_{x\to2^+}f(x)=3(2)-1=5$ Since left and right limits are different, the limit does not exist. Therefore $f$ is not continuous at $\boxed{x=2}$. ---

Continuity and Graph Behaviour

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Why Interval Continuity Matters

These theorems fail easily if continuity on the whole required interval is not present. ---

Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is the correct condition for continuity of a function $f$ on the closed interval $[a,b]$?" options=["$f$ is continuous only on $(a,b)$","$f$ is continuous on $(a,b)$ and $\lim_{x\to a^+}f(x)=f(a),\\ \lim_{x\to b^-}f(x)=f(b)$","$f$ is differentiable on $[a,b]$","$f$ is continuous at $a$ and $b$ only"] answer="B" hint="Endpoints of a closed interval require one-sided continuity." solution="For continuity on the closed interval $[a,b]$, the function must be continuous at every interior point of $(a,b)$, right-continuous at $a$, and left-continuous at $b$. So the correct condition is $\qquad \lim_{x\to a^+}f(x)=f(a)$ and $\qquad \lim_{x\to b^-}f(x)=f(b)$, along with continuity on $(a,b)$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Let $f(x)=\dfrac{1}{x-3}$. How many maximal open intervals of continuity does $f$ have on the real line?" answer="2" hint="Find where the function is undefined." solution="The function $\qquad f(x)=\dfrac{1}{x-3}$ is undefined at $\qquad x=3$. A rational function is continuous wherever it is defined. So $f$ is continuous on the intervals $\qquad (-\infty,3)$ and $\qquad (3,\infty)$. These are the maximal open intervals of continuity. Hence the number of such intervals is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["A polynomial is continuous on all real numbers","A rational function is continuous at every real number","For continuity on $[a,b]$, one-sided limits at endpoints are sufficient together with continuity on $(a,b)$","If $f$ is undefined at a point of an interval, then it cannot be continuous on that whole interval"] answer="A,C,D" hint="Think about domain restrictions and endpoint conditions." solution="1. True. Polynomials are continuous everywhere on $\mathbb{R}$.

False. A rational function can fail to be defined where its denominator is zero.

True. This is exactly the definition of continuity on a closed interval.

True. Continuity on an interval requires the function to be defined at all relevant points.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Find the value of $k$ for which the function $\qquad f(x)= \begin{cases} x^2+k, & x<1 \\ 3, & x=1 \\ 2x+1, & x>1 \end{cases}$ is continuous at $x=1$." answer="$k=2$" hint="Match left-hand limit, right-hand limit, and function value at $x=1$." solution="For continuity at $x=1$, we need $\qquad \lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$. Now $\qquad f(1)=3$ Left-hand limit: $\qquad \lim_{x\to1^-}(x^2+k)=1+k$ Right-hand limit: $\qquad \lim_{x\to1^+}(2x+1)=2(1)+1=3$ So for continuity we need $\qquad 1+k=3$ Hence $\qquad k=2$ Therefore the required value is $\boxed{2}$." ::: ---

Summary

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Part 3: Continuity of piecewise functions

Continuity of Piecewise Functions

Overview

Piecewise functions are functions defined by different formulas on different parts of the domain. The main difficulty is not evaluating them away from the joining point, but checking whether the different pieces fit together smoothly. In CMI-style questions, this topic tests whether you can use left-hand limit, right-hand limit, and function value correctly and without confusion. ---

Learning Objectives

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Core Idea

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Left-Hand and Right-Hand Limits

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Standard Procedure

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Types of Failure

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Minimal Worked Examples

Example 1 Check continuity at $x=1$ for $\qquad f(x)= \begin{cases} x+1, & x<1 \\ 2, & x=1 \\ x^2, & x>1 \end{cases}$ Left-hand limit: $\qquad \lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}(x+1)=2$ Right-hand limit: $\qquad \lim_{x\to 1^+} f(x) = \lim_{x\to 1^+}(x^2)=1$ Since $2\ne1$, the two-sided limit does not exist. Hence the function is not continuous at $x=1$. --- Example 2 Find $k$ such that $\qquad f(x)= \begin{cases} x^2+k, & x<2 \\ 6, & x=2 \\ 3x, & x>2 \end{cases}$ is continuous at $x=2$. For continuity, we need $\qquad \lim_{x\to 2^-} f(x)=f(2)=\lim_{x\to 2^+} f(x)$ Right-hand side: $\qquad \lim_{x\to 2^+}3x=6$ This matches $f(2)=6$. Left-hand side: $\qquad \lim_{x\to 2^-}(x^2+k)=4+k$ For continuity, $\qquad 4+k=6$ $\qquad k=2$ So the required value is $\boxed{2}$. ---

Standard Forms That Appear Often

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Removable Discontinuity

Classic example: $\qquad f(x)= \begin{cases} \dfrac{x^2-1}{x-1}, & x\ne1 \\ k, & x=1 \end{cases}$ For $x\ne1$, $\qquad \dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1$ So, $\qquad \lim_{x\to1} f(x)=2$ Thus continuity at $x=1$ requires $\boxed{k=2}$. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="For the function $f(x)=\begin{cases} x+2, & x<1 \\ 3, & x=1 \\ x^2+1, & x>1 \end{cases}$, which statement is correct?" options=["The function is continuous at $x=1$","The left-hand and right-hand limits are equal but not equal to $f(1)$","The left-hand and right-hand limits are unequal","The function is undefined at $x=1$"] answer="B" hint="Compute the left-hand limit, right-hand limit, and value at $x=1$ separately." solution="Left-hand limit: $\qquad \lim_{x\to1^-}(x+2)=3$ Right-hand limit: $\qquad \lim_{x\to1^+}(x^2+1)=2$ Wait carefully: these are not equal. So the two-sided limit does not exist. Hence the correct statement is that the left-hand and right-hand limits are unequal. Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find the value of $k$ so that the function $f(x)=\begin{cases} x^2+k, & x<2 \\ 6, & x=2 \\ 3x, & x>2 \end{cases}$ is continuous at $x=2$." answer="2" hint="Match LHL, RHL, and $f(2)$." solution="For continuity at $x=2$, we need $\qquad \lim_{x\to2^-}f(x)=f(2)=\lim_{x\to2^+}f(x)$ Now, $\qquad f(2)=6$ Right-hand limit: $\qquad \lim_{x\to2^+}3x=6$ Left-hand limit: $\qquad \lim_{x\to2^-}(x^2+k)=4+k$ For continuity, $\qquad 4+k=6$ So, $\qquad k=2$ Therefore the answer is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$, then $f$ is continuous at $x=a$","A function can be continuous at $x=a$ even if $f(a)$ is not defined","If the left-hand and right-hand limits at $x=a$ are unequal, then $f$ is not continuous at $x=a$","For piecewise functions, checking only $\lim_{x\to a}f(x)$ is always enough"] answer="A,C" hint="Use the definition of continuity carefully." solution="1. True. This is exactly the continuity condition.

False. If $f(a)$ is not defined, continuity at $a$ is impossible.

True. If LHL and RHL are unequal, the two-sided limit does not exist, so continuity fails.

False. In piecewise questions, one usually computes one-sided limits separately and must also compare with $f(a)$.

Hence the correct answer is $\boxed{A,C}$." ::: :::question type="SUB" question="Find the value of $k$ such that $f(x)=\begin{cases} \dfrac{x^2-1}{x-1}, & x\ne1 \\ k, & x=1 \end{cases}$ is continuous at $x=1$." answer="$k=2$" hint="Simplify the rational expression for $x\ne1$." solution="For $x\ne1$, $\qquad \dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1$ So, $\qquad \lim_{x\to1}f(x)=\lim_{x\to1}(x+1)=2$ For continuity at $x=1$, we need $\qquad f(1)=\lim_{x\to1}f(x)$ Thus, $\qquad k=2$ Therefore the required value is $\boxed{2}$." ::: ---

Summary

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Part 4: Intermediate value reasoning

Intermediate Value Reasoning

Overview

Intermediate value reasoning is one of the most important ways to prove that an equation has a real solution without actually solving it. In CMI-style questions, this topic is usually tested through continuity, sign change, and correct use of the Intermediate Value Theorem. The key skill is to show that a continuous function takes two values on opposite sides of a target value, so it must take that target value somewhere in between. ---

Learning Objectives

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Core Idea

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Intermediate Value Theorem

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Why This Works So Often for Polynomials

This is why many exam questions ask you to test a polynomial at nearby integers. ---

Standard Method

For root-existence questions, the target value is usually $0$. ---

Sign Change Reasoning

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Root Between Consecutive Integers

This is exactly the style used in many short-answer exam problems. ---

Intermediate Values Other Than Zero

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Minimal Worked Examples

Example 1 Show that the equation $\qquad x^3-x-1=0$ has a real root between $1$ and $2$. Let $\qquad f(x)=x^3-x-1$ Since $f$ is a polynomial, it is continuous everywhere. Now, $\qquad f(1)=1-1-1=-1$ $\qquad f(2)=8-2-1=5$ Since $f(1)<0<f(2)$, there exists some $c\in(1,2)$ such that $\qquad f(c)=0$ So the equation has a real root in $(1,2)$. --- Example 2 Show that a continuous function with $f(2)=3$ and $f(5)=9$ must take the value $6$ somewhere in $(2,5)$. Since $f$ is continuous on $[2,5]$ and $\qquad 3<6<9$ IVT gives some $c\in(2,5)$ such that $\qquad f(c)=6$ ---

What IVT Does Not Say

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When IVT Cannot Be Used Directly

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Common Question Patterns

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which theorem is most directly used to prove that a continuous function with $f(a)<0<f(b)$ has a root in $(a,b)$?" options=["Mean Value Theorem","Intermediate Value Theorem","Factor Theorem","Rolle's Theorem"] answer="B" hint="Think about value crossing, not derivative information." solution="The statement that a continuous function taking opposite signs at two points must be zero somewhere in between is a direct application of the Intermediate Value Theorem. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the least non-negative integer $n$ such that $x^3-2$ has a real root in $(n,n+1)$." answer="1" hint="Check the values at consecutive integers." solution="Let $f(x)=x^3-2$. This is a polynomial, so it is continuous everywhere. Now, $\qquad f(0)=-2$ $\qquad f(1)=-1$ $\qquad f(2)=6$ There is no sign change on $(0,1)$, but $\qquad f(1)<0<f(2)$ So by IVT, there is a root in $(1,2)$. Hence the least non-negative integer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Every polynomial is continuous on $\mathbb{R}$","If a continuous function changes sign on an interval, then it has at least one root there","IVT always gives the exact location of the root","A continuous function with $f(a)=2$ and $f(b)=7$ must take every value between $2$ and $7$"] answer="A,B,D" hint="IVT proves existence, not exact position." solution="1. True. Every polynomial is continuous on all real numbers.

True. This is the root-existence form of IVT.

False. IVT gives existence, not the exact location.

True. That is exactly what IVT says.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Show that the equation $x^5-x-1=0$ has a real root in $(1,2)$." answer="Since the polynomial is continuous and changes sign between $1$ and $2$, IVT guarantees a root in $(1,2)$" hint="Evaluate the polynomial at $x=1$ and $x=2$." solution="Let $\qquad f(x)=x^5-x-1$ Since $f$ is a polynomial, it is continuous on $[1,2]$. Now, $\qquad f(1)=1-1-1=-1$ and $\qquad f(2)=32-2-1=29$ Thus $\qquad f(1)<0<f(2)$ So by the Intermediate Value Theorem, there exists some $c\in(1,2)$ such that $\qquad f(c)=0$ Hence the equation has a real root in $(1,2)$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="For what value of $a$ is the function $f(x)$ continuous everywhere?

" options=["$a=0$", "$a=1$", "$a=3$", "$a=-3$"] answer="$a=3$" hint="For continuity at $x=0$, the left-hand limit, right-hand limit, and function value must all be equal." solution="For $f(x)$ to be continuous everywhere, it must be continuous at $x=0$.
We need $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

First, calculate $f(0)$:
$f(0) = 2(0) + 3 = 3$.

Next, calculate the right-hand limit:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x+3) = 2(0)+3 = 3$.

Finally, calculate the left-hand limit:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(ax)}{x}$.
Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$, we can write:
$\lim_{x \to 0^-} \frac{\sin(ax)}{x} = \lim_{x \to 0^-} \left( a \cdot \frac{\sin(ax)}{ax} \right) = a \cdot 1 = a$.

For continuity at $x=0$, we must have $a = 3 = 3$.
Therefore, $a=3$.

The correct option is $a=3$."
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:::question type="NAT" question="Consider the function $g(x) = \frac{x^2 - 4}{x-2}$. If $g(x)$ is redefined at $x=2$ to make it continuous, what value should $g(2)$ be assigned?" answer="4" hint="Simplify the expression for $g(x)$ for $x \neq 2$ to find the limit as $x \to 2$." solution="The function $g(x) = \frac{x^2 - 4}{x-2}$ is undefined at $x=2$.
For $x \neq 2$, we can simplify the expression:
$g(x) = \frac{(x-2)(x+2)}{x-2} = x+2$.

To make $g(x)$ continuous at $x=2$, we need to define $g(2)$ such that $g(2) = \lim_{x \to 2} g(x)$.
$\lim_{x \to 2} g(x) = \lim_{x \to 2} (x+2) = 2+2 = 4$.

Thus, to make $g(x)$ continuous, $g(2)$ should be assigned the value 4."
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:::question type="MCQ" question="Given a continuous function $f(x)$ on the interval $[1, 5]$ such that $f(1)=-3$ and $f(5)=7$. Which of the following statements is guaranteed by the Intermediate Value Theorem?" options=["There exists $c \in (1, 5)$ such that $f(c)=0$.", "There exists $c \in (1, 5)$ such that $f(c)=10$.", "The function $f(x)$ is increasing on $[1, 5]$.", "The function $f(x)$ has a maximum value of 7 and a minimum value of -3 on $[1, 5]$."] answer="There exists $c \in (1, 5)$ such that $f(c)=0$." hint="The IVT guarantees that a continuous function takes on every value between $f(a)$ and $f(b)$." solution="The Intermediate Value Theorem (IVT) states that if $f$ is continuous on $[a,b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c \in [a,b]$ such that $f(c)=k$.

Here, $a=1$, $b=5$, $f(a)=-3$, and $f(b)=7$.
The range of values that $f(x)$ is guaranteed to take on is $[-3, 7]$.

* Option 1: There exists $c \in (1, 5)$ such that $f(c)=0$.
Since $0$ is between $-3$ and $7$ (i.e., $f(1) < 0 < f(5)$), the IVT guarantees that there exists at least one $c \in (1, 5)$ such that $f(c)=0$. This statement is true.

* Option 2: There exists $c \in (1, 5)$ such that $f(c)=10$.
Since $10$ is not between $-3$ and $7$, the IVT does not guarantee that $f(c)=10$ for any $c \in (1, 5)$. (The function might reach 10, but it's not guaranteed).

* Option 3: The function $f(x)$ is increasing on $[1, 5]$.
The IVT does not imply monotonicity. A continuous function can oscillate while still passing through all intermediate values.

* Option 4: The function $f(x)$ has a maximum value of 7 and a minimum value of -3 on $[1, 5]$.
While the Extreme Value Theorem (another theorem about continuous functions on closed intervals) guarantees that $f$ attains a maximum and minimum value, it does not state that these values must be $f(1)$ and $f(5)$. For example, $f(x)$ could go down to $-5$ or up to $10$ at some point within $(1,5)$ before returning to $f(5)=7$. The values $f(1)$ and $f(5)$ are just the function values at the endpoints, not necessarily the global extrema.

Therefore, the only statement guaranteed by the IVT is that there exists $c \in (1, 5)$ such that $f(c)=0$."
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What's Next?