Limits

This chapter rigorously introduces the concept of limits, a fundamental pillar of calculus essential for subsequent topics such as continuity, differentiation, and integration. A thorough understanding of the various limit evaluation techniques presented herein is critical, as these concepts frequently appear in CMI examinations, often requiring precise application to complex functions.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Algebraic limits | | 2 | One-sided limits | | 3 | Trigonometric limits | | 4 | Limits involving logarithm and exponential | | 5 | Infinite limits |

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We begin with Algebraic limits.

Part 1: Algebraic limits

Algebraic Limits

Overview

Algebraic limits are limits that can be evaluated mainly using algebraic manipulation instead of advanced theorems. In CMI-style problems, the real test is usually not the final substitution, but identifying the right simplification: factorisation, cancellation, rationalisation, or rewriting expressions into a standard form. ---

Learning Objectives

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Core Idea

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First Check: Direct Substitution

Example $\qquad \lim_{x\to 2}(3x^2-5x+1)=3(2)^2-5(2)+1=12-10+1=3$ ---

The Main Warning Sign

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Method 1: Factor and Cancel

Example 1 Evaluate $\qquad \lim_{x\to 3}\dfrac{x^2-9}{x-3}$ Factor numerator: $\qquad x^2-9=(x-3)(x+3)$ So, $\qquad \lim_{x\to 3}\dfrac{x^2-9}{x-3} = \lim_{x\to 3}(x+3)=6$ Hence the limit is $\boxed{6}$. ---

Method 2: Rationalise Using Conjugates

Example 2 Evaluate $\qquad \lim_{x\to 4}\dfrac{\sqrt{x}-2}{x-4}$ Multiply numerator and denominator by $\sqrt{x}+2$: $\qquad \dfrac{\sqrt{x}-2}{x-4}\cdot \dfrac{\sqrt{x}+2}{\sqrt{x}+2} = \dfrac{x-4}{(x-4)(\sqrt{x}+2)} = \dfrac{1}{\sqrt{x}+2}$ Now substitute $x=4$: $\qquad \lim_{x\to 4}\dfrac{1}{\sqrt{x}+2} = \dfrac{1}{4}$ So the limit is $\boxed{\dfrac{1}{4}}$. ---

Method 3: Common Algebraic Patterns

These are best understood by algebra, not memorised blindly. ---

Difference of Powers

Example 3 Evaluate $\qquad \lim_{x\to 2}\dfrac{x^3-8}{x-2}$ Factor: $\qquad x^3-8=(x-2)(x^2+2x+4)$ So, $\qquad \lim_{x\to 2}\dfrac{x^3-8}{x-2} = \lim_{x\to 2}(x^2+2x+4)=4+4+4=12$ Hence the limit is $\boxed{12}$. ---

Rational Expressions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\lim\limits_{x\to 2}\dfrac{x^2-4}{x-2}$ is" options=["$2$","$4$","$6$","Does not exist"] answer="B" hint="Factor the numerator." solution="We factor: $\qquad x^2-4=(x-2)(x+2)$ So, $\qquad \dfrac{x^2-4}{x-2}=x+2 \quad \text{for } x\ne2$ Hence, $\qquad \lim_{x\to2}\dfrac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4$ Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $\lim\limits_{x\to 9}\dfrac{\sqrt{x}-3}{x-9}$." answer="0.1667" hint="Use the conjugate of the numerator." solution="Multiply numerator and denominator by $\sqrt{x}+3$: $\qquad \dfrac{\sqrt{x}-3}{x-9}\cdot \dfrac{\sqrt{x}+3}{\sqrt{x}+3} = \dfrac{x-9}{(x-9)(\sqrt{x}+3)} = \dfrac{1}{\sqrt{x}+3}$ Now take the limit: $\qquad \lim_{x\to9}\dfrac{1}{\sqrt{x}+3}=\dfrac{1}{6}$ So the answer is $\boxed{\dfrac{1}{6}}$. In decimal form, $\qquad \dfrac{1}{6}\approx 0.1667$ ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If direct substitution gives a nonzero denominator, the limit of a rational function can be found by substitution","If substitution gives $\dfrac{0}{0}$, the limit does not exist","For $a>0$, $\lim\limits_{x\to a}\dfrac{\sqrt{x}-\sqrt{a}}{x-a}=\dfrac{1}{2\sqrt{a}}$","The identity $x^2-a^2=(x-a)(x+a)$ is useful in algebraic limits"] answer="A,C,D" hint="Think about standard algebraic simplifications." solution="1. True. Rational functions are continuous where the denominator is nonzero.

False. The form $\dfrac{0}{0}$ is indeterminate, not impossible; simplification is needed.

True. This follows by rationalising the numerator.

True. Difference of squares is one of the most important factorisations in limit problems.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Evaluate $\lim\limits_{x\to 3}\dfrac{x^3-27}{x-3}$." answer="$27$" hint="Use difference of cubes." solution="We use $\qquad x^3-27=(x-3)(x^2+3x+9)$ So, $\qquad \dfrac{x^3-27}{x-3}=x^2+3x+9 \quad \text{for } x\ne3$ Now take the limit: $\qquad \lim_{x\to3}(x^2+3x+9)=9+9+9=27$ Therefore the answer is $\boxed{27}$." ::: ---

Summary

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Part 2: One-sided limits

One-Sided Limits

Overview

A one-sided limit studies the behaviour of a function as the input approaches a point from only one side. This topic is central in calculus because many functions behave differently from the left and from the right, especially piecewise functions, modulus-type expressions, rational functions with domain restrictions, and step-like graphs. In CMI-style questions, the real test is usually careful local analysis near the point, not blind substitution. ---

Learning Objectives

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Core Idea

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One-Sided Limits and Continuity

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Standard Situations

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How to Compute One-Sided Limits

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Minimal Worked Examples

Example 1 Let $\qquad f(x)=\begin{cases} x+1, & x<2 \\ 5-x, & x\ge 2 \end{cases}$ Find the one-sided limits at $x=2$. From the left, use $x+1$: $\qquad \lim_{x\to 2^-} f(x)=2+1=3$ From the right, use $5-x$: $\qquad \lim_{x\to 2^+} f(x)=5-2=3$ So both one-sided limits are equal, and hence $\qquad \lim_{x\to 2} f(x)=3$ --- Example 2 Find the one-sided limits of $\qquad \dfrac{|x|}{x}$ at $x=0$. If $x<0$, then $|x|=-x$, so $\qquad \dfrac{|x|}{x}=\dfrac{-x}{x}=-1$ Hence, $\qquad \lim_{x\to 0^-}\dfrac{|x|}{x}=-1$ If $x>0$, then $|x|=x$, so $\qquad \dfrac{|x|}{x}=\dfrac{x}{x}=1$ Hence, $\qquad \lim_{x\to 0^+}\dfrac{|x|}{x}=1$ Since the two one-sided limits are different, the two-sided limit does not exist. ---

Common Patterns

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One-Sided Limits and Graph Reading

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Important Consequences

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Let $f(x)=\begin{cases}2x+1, & x<1\x+2, & x\ge1\end{cases}$. Then $\lim_{x\to1^-}f(x)$ is" options=["$1$","$2$","$3$","$4$"] answer="C" hint="Use the formula valid for $x<1$." solution="For the left-hand limit at $x=1$, we use the branch valid for $x<1$, namely $f(x)=2x+1$. So, $\qquad \lim_{x\to1^-}f(x)=2(1)+1=3$ Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $\lim_{x\to0^+}\dfrac{|x|}{x}$." answer="1" hint="For $x>0$, $|x|=x$." solution="As $x\to0^+$, we have $x>0$, so $\qquad |x|=x$ Hence, $\qquad \dfrac{|x|}{x}=\dfrac{x}{x}=1$ Therefore, $\qquad \lim_{x\to0^+}\dfrac{|x|}{x}=1$ So the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist and are equal, then $\lim_{x\to a}f(x)$ exists","If $f(a)$ is defined, then $\lim_{x\to a}f(x)$ must exist","A two-sided limit can exist even if $f(a)$ is different from the limit value","If the left-hand and right-hand limits are unequal, then the two-sided limit does not exist"] answer="A,C,D" hint="Separate the limit from the actual function value." solution="1. True. Equality of the two one-sided limits gives existence of the two-sided limit.

False. Merely defining $f(a)$ does not force the limit to exist.

True. The limit depends on nearby behaviour, not necessarily on the value at the point.

True. Unequal one-sided limits imply that the two-sided limit does not exist.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="For the function $f(x)=\begin{cases}x^2, & x<1\\2x-1, & x\ge1\end{cases}$, examine the limit and continuity at $x=1$." answer="The limit exists and equals $1$; the function is continuous at $x=1$." hint="Compute both one-sided limits and compare with $f(1)$." solution="For $x<1$, $f(x)=x^2$. Hence $\qquad \lim_{x\to1^-}f(x)=1^2=1$ For $x\ge1$, $f(x)=2x-1$. Hence $\qquad \lim_{x\to1^+}f(x)=2(1)-1=1$ Since the two one-sided limits are equal, we get $\qquad \lim_{x\to1}f(x)=1$ Now, $\qquad f(1)=2(1)-1=1$ So, $\qquad \lim_{x\to1}f(x)=f(1)$ Therefore the function is continuous at $x=1$, and the limit equals $\boxed{1}$." ::: ---

Summary

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Part 3: Trigonometric limits

Trigonometric Limits

Overview

Trigonometric limits are among the most important standard limits in calculus. They appear directly in limit computation and indirectly in differentiation, continuity, small-angle approximation, and local behaviour of functions. In CMI-style questions, the real test is not memorising one formula, but knowing how to rewrite an expression into a standard form. ---

Learning Objectives

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Core Standard Limits

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Why $\dfrac{\sin x}{x}$ Matters

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Derived Standard Forms

These come from substitution. For example, $\qquad \dfrac{\sin(ax)}{x} = a \cdot \dfrac{\sin(ax)}{ax}$ and as $x\to 0$, we have $ax\to 0$. ---

Essential Identities for Limit Problems

These identities help convert an unfamiliar expression into a standard limit. ---

Standard Techniques

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Minimal Worked Examples

Example 1 Evaluate $\qquad \lim_{x\to 0}\dfrac{\sin 5x}{x}$ Write $\qquad \dfrac{\sin 5x}{x} = 5\cdot \dfrac{\sin 5x}{5x}$ As $x\to 0$, we get $\qquad \dfrac{\sin 5x}{5x}\to 1$ So the limit is $\qquad 5$ --- Example 2 Evaluate $\qquad \lim_{x\to 0}\dfrac{1-\cos x}{x^2}$ Use $\qquad 1-\cos x = 2\sin^2\left(\dfrac{x}{2}\right)$ Then $\qquad \dfrac{1-\cos x}{x^2} = \dfrac{2\sin^2(x/2)}{x^2} = \dfrac{1}{2}\left(\dfrac{\sin(x/2)}{x/2}\right)^2$ As $x\to 0$, $\qquad \dfrac{\sin(x/2)}{x/2}\to 1$ Hence the limit is $\qquad \dfrac{1}{2}$ ---

Small-Angle Approximation

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Common Traps

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\lim_{x\to 0}\dfrac{\sin 7x}{x}$ is" options=["$1$","$7$","$\dfrac{1}{7}$","$0$"] answer="B" hint="Create the form $\dfrac{\sin u}{u}$." solution="Write $\qquad \dfrac{\sin 7x}{x}=7\cdot \dfrac{\sin 7x}{7x}$ As $x\to 0$, $\qquad \dfrac{\sin 7x}{7x}\to 1$ Hence the limit is $\qquad 7$ So the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $\lim_{x\to 0}\dfrac{1-\cos 2x}{x^2}$." answer="2" hint="Use the standard limit for $\dfrac{1-\cos(ax)}{x^2}$." solution="Using the identity $\qquad 1-\cos 2x = 2\sin^2 x$ we get $\qquad \dfrac{1-\cos 2x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2$ As $x\to 0$, $\qquad \dfrac{\sin x}{x}\to 1$ Therefore the limit is $\qquad 2\cdot 1^2 = 2$ Hence the answer is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\lim_{x\to 0}\dfrac{\tan x}{x}=1$","$\lim_{x\to 0}\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}$","$\lim_{x\to 0}\dfrac{\sin x}{x}=0$","$\lim_{x\to 0}\dfrac{\cos x-1}{x^2}=-\dfrac{1}{2}$"] answer="A,B,D" hint="Recall the standard trigonometric limits exactly." solution="1. True.

True.

False, because $\lim_{x\to 0}\dfrac{\sin x}{x}=1$.

True, since

$\qquad \dfrac{\cos x-1}{x^2}=-\dfrac{1-\cos x}{x^2}\to -\dfrac{1}{2}$ Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Evaluate $\lim_{x\to 0}\dfrac{\sin 3x}{\tan 2x}$." answer="$\dfrac{3}{2}$" hint="Break it into two standard factors." solution="Write $\qquad \dfrac{\sin 3x}{\tan 2x} = \dfrac{\sin 3x}{3x}\cdot \dfrac{3x}{2x}\cdot \dfrac{2x}{\tan 2x}$ So, $\qquad \dfrac{\sin 3x}{\tan 2x} = \left(\dfrac{\sin 3x}{3x}\right)\cdot \dfrac{3}{2}\cdot \left(\dfrac{2x}{\tan 2x}\right)$ As $x\to 0$, $\qquad \dfrac{\sin 3x}{3x}\to 1$ and $\qquad \dfrac{2x}{\tan 2x}\to 1$ Therefore the limit is $\qquad 1\cdot \dfrac{3}{2}\cdot 1 = \dfrac{3}{2}$ Hence the answer is $\boxed{\dfrac{3}{2}}$." ::: ---

Summary

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Part 4: Limits involving logarithm and exponential

Limits Involving Logarithm and Exponential

Overview

Limits involving logarithms and exponentials are among the most important small-quantity limits in calculus. In CMI-style problems, these are rarely tested as isolated formulas only; they appear inside substitutions, quotients, compositions, asymptotic comparisons, and even nested limits. The key skill is to identify the small parameter and then replace complicated expressions by their correct leading-order behavior. ---

Learning Objectives

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Core Idea

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Fundamental Standard Limits

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First-Order Equivalents

These can be combined. For example, as $x\to0$, $\qquad \dfrac{\ln(1+2x)}{e^{3x}-1} \sim \dfrac{2x}{3x}=\dfrac{2}{3}$ ---

Second-Order Expansions

These are used when first-order terms cancel. ---

Growth Comparison at Infinity

This is exactly the type of reasoning needed in limits such as $\qquad \lim_{x\to\infty}\dfrac{x^4+x^6}{e^x-1-x^2}$ because the denominator is dominated by $e^x$. ---

Parameter and Composition Method

Example $\qquad \lim_{x\to0}\dfrac{\ln(1+x^2)}{x^2}=1$ because with $u=x^2$ we have $\ln(1+u)\sim u$. ---

Minimal Worked Examples

Example 1 Evaluate $\qquad \lim_{x\to0}\dfrac{\ln(1+3x)}{x}$ Using $\ln(1+u)\sim u$ with $u=3x$, $\qquad \dfrac{\ln(1+3x)}{x} \sim \dfrac{3x}{x}=3$ So the limit is $\boxed{3}$. --- Example 2 Evaluate $\qquad \lim_{x\to0}\dfrac{\ln(1+2x)}{e^{3x}-1}$ Use first-order equivalents: $\qquad \ln(1+2x)\sim2x,\qquad e^{3x}-1\sim3x$ Hence, $\qquad \lim_{x\to0}\dfrac{\ln(1+2x)}{e^{3x}-1}=\dfrac{2}{3}$ So the limit is $\boxed{\dfrac{2}{3}}$. --- Example 3 Evaluate $\qquad \lim_{x\to0}\dfrac{e^x-1-\ln(1+x)}{x^2}$ Using second-order expansions, $\qquad e^x-1=x+\dfrac{x^2}{2}+o(x^2)$ $\qquad \ln(1+x)=x-\dfrac{x^2}{2}+o(x^2)$ So, $\qquad e^x-1-\ln(1+x)=x+\dfrac{x^2}{2}-x+\dfrac{x^2}{2}+o(x^2)=x^2+o(x^2)$ Therefore, $\qquad \lim_{x\to0}\dfrac{e^x-1-\ln(1+x)}{x^2}=1$ Hence the limit is $\boxed{1}$. ---

PYQ-Relevant Insight

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\lim\limits_{x\to0}\dfrac{\ln(1+2x)}{x}$ is" options=["$1$","$2$","$\dfrac{1}{2}$","$0$"] answer="B" hint="Use $\ln(1+u)\sim u$." solution="As $x\to0$, we have $\qquad \ln(1+2x)\sim 2x$. Therefore, $\qquad \lim_{x\to0}\dfrac{\ln(1+2x)}{x}=2$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $\lim\limits_{x\to0}\dfrac{e^{3x}-1}{x}$." answer="3" hint="Use $e^u-1\sim u$ with $u=3x$." solution="As $x\to0$, $\qquad e^{3x}-1\sim 3x$. Hence, $\qquad \lim_{x\to0}\dfrac{e^{3x}-1}{x}=3$. Therefore the answer is $\boxed{3}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\lim\limits_{x\to0}\dfrac{\ln(1+x)}{x}=1$","$\lim\limits_{x\to0}\dfrac{e^x-1}{x}=1$","$\lim\limits_{x\to\infty}\dfrac{x^3}{e^x}$ is finite","$\lim\limits_{x\to\infty}\dfrac{\ln x}{x}=1$"] answer="A,B,C" hint="Use standard limits and growth comparison." solution="1. True. This is a standard logarithmic limit.

True. This is a standard exponential limit.

True. Since exponential growth dominates polynomial growth, $\dfrac{x^3}{e^x}\to0$, which is finite.

False. Since $\ln x$ grows slower than $x$, $\dfrac{\ln x}{x}\to0$.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Evaluate $\lim\limits_{x\to0}\dfrac{e^x-1-\ln(1+x)}{x^2}$." answer="$1$" hint="Use second-order expansions." solution="As $x\to0$, $\qquad e^x-1=x+\dfrac{x^2}{2}+o(x^2)$ and $\qquad \ln(1+x)=x-\dfrac{x^2}{2}+o(x^2)$. Therefore, $\qquad e^x-1-\ln(1+x)=x+\dfrac{x^2}{2}-x+\dfrac{x^2}{2}+o(x^2)=x^2+o(x^2)$. Hence, $\qquad \lim_{x\to0}\dfrac{e^x-1-\ln(1+x)}{x^2}=1$. So the answer is $\boxed{1}$." ::: ---

Summary

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Part 5: Infinite limits

Infinite Limits

Overview

Infinite limits describe what happens when the values of a function become arbitrarily large in magnitude near a point. In calculus, this topic is closely tied to vertical asymptotes, sign analysis, and one-sided limits. In CMI-style questions, the real test is often not symbol pushing, but understanding from which side the function approaches the point and whether it tends to $+\infty$ or $-\infty$. ---

Learning Objectives

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Core Idea

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Infinite Limit and Vertical Asymptote

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One-Sided Limits Matter

For example, for $\qquad f(x)=\dfrac{1}{x}$ we have $\qquad \lim_{x\to 0^-}\dfrac{1}{x}=-\infty$ and $\qquad \lim_{x\to 0^+}\dfrac{1}{x}=+\infty$ So the two-sided limit at $0$ does not exist, even though each one-sided limit is infinite. ---

Standard Algebraic Situations

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Sign Rules Near a Point

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Minimal Worked Examples

Example 1 Find $\qquad \lim_{x\to 2}\dfrac{1}{(x-2)^2}$ Since $(x-2)^2>0$ for $x\ne 2$ and becomes very small positive near $x=2$, $\qquad \dfrac{1}{(x-2)^2}\to +\infty$ Therefore, $\qquad \lim_{x\to 2}\dfrac{1}{(x-2)^2}=+\infty$ --- Example 2 Find $\qquad \lim_{x\to 3^-}\dfrac{1}{x-3}$ As $x\to 3^-$, the quantity $x-3$ is negative and very close to $0$. So, $\qquad \dfrac{1}{x-3}\to -\infty$ Hence, $\qquad \lim_{x\to 3^-}\dfrac{1}{x-3}=-\infty$ ---

Rational Functions

If both numerator and denominator vanish, then more analysis is needed; the limit may be finite, infinite, or fail to exist. ---

Even and Odd Multiplicity Idea

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Infinite Limit Does Not Mean Ordinary Limit Exists

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Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\lim_{x\to 1}\dfrac{1}{(x-1)^2}$ is" options=["$0$","Does not exist","$+\infty$","$-\infty$"] answer="C" hint="Square in the denominator controls the sign." solution="As $x\to 1$, the quantity $(x-1)^2$ becomes very small and remains positive. Therefore $\qquad \dfrac{1}{(x-1)^2}$ becomes arbitrarily large positive. So $\qquad \lim_{x\to 1}\dfrac{1}{(x-1)^2}=+\infty$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $f(x)=\dfrac{1}{x-4}$, then find the value of $\lim_{x\to 4^-} f(x)$ in the form $+\infty$ or $-\infty$." answer="-infinity" hint="Approach $4$ from the left." solution="When $x\to 4^-$, we have $\qquad x-4<0$ and it becomes very close to $0$. Therefore $\qquad \dfrac{1}{x-4}$ becomes very large negative. So $\qquad \lim_{x\to 4^-}\dfrac{1}{x-4}=-\infty$. Hence the answer is $\boxed{-\infty}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $\lim_{x\to a^-}f(x)=+\infty$ and $\lim_{x\to a^+}f(x)=+\infty$, then $x=a$ is a vertical asymptote","If $\lim_{x\to a^-}f(x)=-\infty$ and $\lim_{x\to a^+}f(x)=+\infty$, then the two-sided limit at $a$ exists","For $\dfrac{1}{(x-a)^2}$, the left-hand and right-hand infinite limits have the same sign","For $\dfrac{1}{x-a}$, the left-hand and right-hand infinite limits always have the same sign"] answer="A,C" hint="Check one-sided behavior carefully." solution="1. True. If the function becomes unbounded near $x=a$, then $x=a$ is a vertical asymptote.

False. If the left and right behaviors differ, the two-sided limit does not exist.

True. Because $(x-a)^2$ is positive on both sides of $a$, both sides have the same sign.

False. For $\dfrac{1}{x-a}$, the sign changes across $a$.

Hence the correct answer is $\boxed{A,C}$." ::: :::question type="SUB" question="Find $\lim_{x\to 2^-}\dfrac{3}{x-2}$ and $\lim_{x\to 2^+}\dfrac{3}{x-2}$. Hence comment on $\lim_{x\to 2}\dfrac{3}{x-2}$." answer="Left-hand limit is $-\infty$, right-hand limit is $+\infty$, so the two-sided limit does not exist" hint="Check the sign of $x-2$ on each side of $2$." solution="As $x\to 2^-$, we have $\qquad x-2<0$ and very close to $0$. Therefore $\qquad \dfrac{3}{x-2}\to -\infty$. As $x\to 2^+$, we have $\qquad x-2>0$ and very close to $0$. Therefore $\qquad \dfrac{3}{x-2}\to +\infty$. So, $\qquad \lim_{x\to 2^-}\dfrac{3}{x-2}=-\infty$ and $\qquad \lim_{x\to 2^+}\dfrac{3}{x-2}=+\infty$. Since the left-hand and right-hand limits are different, the two-sided limit $\qquad \lim_{x\to 2}\dfrac{3}{x-2}$ does not exist." ::: ---

Summary

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Evaluate $\lim_{x \to -1} \frac{x^3+1}{x^2-1}$." options=["$-\frac{3}{2}$", "$\frac{3}{2}$", "$-1$", "$0$"] answer="$-\frac{3}{2}$" hint="Factorize the numerator and denominator to cancel common terms." solution="The numerator $x^3+1$ factors as $(x+1)(x^2-x+1)$. The denominator $x^2-1$ factors as $(x-1)(x+1)$.
So, $\lim_{x \to -1} \frac{x^3+1}{x^2-1} = \lim_{x \to -1} \frac{(x+1)(x^2-x+1)}{(x-1)(x+1)} = \lim_{x \to -1} \frac{x^2-x+1}{x-1}$.
Substituting $x=-1$, we get $\frac{(-1)^2 - (-1) + 1}{-1-1} = \frac{1+1+1}{-2} = \frac{3}{-2} = -\frac{3}{2}$."
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:::question type="NAT" question="Let $f(x) = \begin{cases} 2x+k & x < 2 \\ x^2-1 & x \ge 2 \end{cases}$. For $\lim_{x \to 2} f(x)$ to exist, what must be the value of $k$?" answer="-1" hint="For a limit to exist at a point, the left-hand limit and the right-hand limit must be equal." solution="For $\lim_{x \to 2} f(x)$ to exist, the left-hand limit must equal the right-hand limit.
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x+k) = 2(2)+k = 4+k$.
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2-1) = 2^2-1 = 4-1 = 3$.
Setting these equal: $4+k = 3 \implies k = 3-4 = -1$."
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:::question type="MCQ" question="Evaluate $\lim_{x \to 0} \frac{\tan(3x)}{x}$." options=["$0$", "$1$", "$3$", "$\infty$"] answer="$3$" hint="Rewrite $\tan(3x)$ in terms of $\sin(3x)$ and $\cos(3x)$, then use the special limit $\lim_{u \to 0} \frac{\sin u}{u}=1$." solution="We can rewrite the expression as:
$\lim_{x \to 0} \frac{\tan(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{x \cos(3x)}$.
To use the special limit $\lim_{u \to 0} \frac{\sin u}{u}=1$, we multiply and divide by 3:
$\lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3}{\cos(3x)}$.
As $x \to 0$, $3x \to 0$, so $\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$.
Also, $\lim_{x \to 0} \cos(3x) = \cos(0) = 1$.
Therefore, the limit is $1 \cdot \frac{3}{1} = 3$."
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:::question type="NAT" question="Evaluate $\lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}$." answer="0.25" hint="This is an indeterminate form $0/0$. Multiply the numerator and denominator by the conjugate of the numerator." solution="This is an indeterminate form $0/0$. We multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+4}+2$:

For $x \ne 0$, we can cancel $x$:

Now, substitute $x=0$:

As a decimal, $\frac{1}{4} = 0.25$."
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What's Next?