Limits and Continuity

This chapter rigorously introduces the concepts of limits and continuity, providing the foundational analytical tools essential for advanced calculus. Mastery of these topics is critical for solving problems involving function behavior, convergence, and forms a prerequisite for subsequent material on differentiation and integration in the CMI examination.

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Chapter Contents

| Topic |

|---|-------| | 1 | Limits | | 2 | Continuity |

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We begin with Limits.

Part 1: Limits

Limits are fundamental to calculus, forming the basis for continuity, derivatives, and integrals. We use limits to analyze the behavior of functions as their input approaches a particular value or infinity.

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Core Concepts

1. Intuitive Definition of a Limit

We say that the limit of $f(x)$ as $x$ approaches $c$ is $L$ , written as $\lim_{x \to c} f(x) = L$ , if $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close to $c$ , but not necessarily equal to $c$ .

Worked Example:

Consider the function $f(x) = \frac{x^2 - 1}{x - 1}$ . We want to find $\lim_{x \to 1} f(x)$ .

Step 1: Observe the function's behavior near $x=1$ .

The function is undefined at $x=1$ . We can factor the numerator.

* * S t e p 2 : * * S i m p l i f y t h e e x p r e s s i o n f o r

f(x) = x+1 \quad \text{for } x \neq 1

x

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x+1) = 2

2

x^2 - 9 = (x-3)(x+3)

* * S t e p 2 : * * S u b s t i t u t e t h e f a c t o r e d f o r m i n t o t h e l i m i t e x p r e s s i o n . >

\lim_{x \to 3} \frac{(x-3)(x+3)}{x-3}

, we can cancel the

\lim_{x \to 3} (x+3) $* * S t e p 4 : * * E v a l u a t e t h e l i m i t b y d i r e c t s u b s t i t u t i o n . >$

3+3 = 6
$f(x)$ |(3x - 1) - 5| < \epsilon $>$

|3x - 6| < \epsilon
$>$

|3(x - 2)| < \epsilon
$>$

3|x - 2| < \epsilon
$|x - c|$ |x - 2| < \frac{\epsilon}{3} $\delta$ |(2x + 3) - 11| < \epsilon $>$

|2x - 8| < \epsilon
$>$

|2(x - 4)| < \epsilon
$>$

2|x - 4| < \epsilon
$|x - c|$ |x - 4| < \frac{\epsilon}{2} $\delta$ \lim_{x \to 2} [2f(x) - 5g(x)]^3 = \left[ \lim_{x \to 2} (2f(x) - 5g(x)) \right]^3 $* * S t e p 2 : * * A p p l y t h e C o n s t a n t M u l t i p l e a n d D i f f e r e n c e L a w s . >$

= \left[ 2 \lim_{x \to 2} f(x) - 5 \lim_{x \to 2} g(x) \right]^3
$* * S t e p 3 : * * S u b s t i t u t e t h e g i v e n l i m i t v a l u e s . >$

= [2(3) - 5(-1)]^3
$>$

= [6 + 5]^3
$>$

= [11]^3
$>$

= 1331
$1331$ \lim_{x \to 0} \frac{f(x) - h(x)}{f(x) \cdot h(x)} = \frac{\lim_{x \to 0} (f(x) - h(x))}{\lim_{x \to 0} (f(x) \cdot h(x))} $* * S t e p 2 : * * A p p l y t h e D i f f e r e n c e L a w t o t h e n u m e r a t o r a n d t h e P r o d u c t L a w t o t h e d e n o m i n a t o r . >$

= \frac{\lim_{x \to 0} f(x) - \lim_{x \to 0} h(x)}{\lim_{x \to 0} f(x) \cdot \lim_{x \to 0} h(x)}
$* * S t e p 3 : * * S u b s t i t u t e t h e g i v e n l i m i t v a l u e s . >$

= \frac{4 - (-2)}{4 \cdot (-2)}
$>$

= \frac{6}{-8}
$>$

= -\frac{3}{4}
$-3/4$ \cos(\pi) + (\pi)^2 $>$

-1 + \pi^2
$\pi^2 - 1$ \lim_{x \to 1} \frac{x^3 + 2x - 5}{x^2 + 1} = \frac{(1)^3 + 2(1) - 5}{(1)^2 + 1} $>$

= \frac{1 + 2 - 5}{1 + 1}
$>$

= \frac{-2}{2}
$>$

= -1
$-1$ (2+h)^2 - 4 = (4 + 4h + h^2) - 4 $>$

= 4h + h^2
$* * S t e p 3 : * * S u b s t i t u t e t h e e x p a n d e d f o r m b a c k i n t o t h e l i m i t a n d f a c t o r . >$

\lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} \frac{h(4 + h)}{h}
$(since$ = \lim_{h \to 0} (4 + h) $* * S t e p 5 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

= 4 + 0 = 4
$4$ x^3 + 8 = (x+2)(x^2 - 2x + 4) $* * S t e p 3 : * * S u b s t i t u t e t h e f a c t o r e d f o r m i n t o t h e l i m i t e x p r e s s i o n . >$

\lim_{x \to -2} \frac{(x+2)(x^2 - 2x + 4)}{x + 2}
$(since$ \lim_{x \to -2} (x^2 - 2x + 4) $* * S t e p 5 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

(-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12
$\frac{0}{0}$ \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} $(a-b)(a+b) = a^2 - b^2$ = \lim_{x \to 0} \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} $>$

= \lim_{x \to 0} \frac{x}{x(\sqrt{x+4} + 2)}
$(since$ = \lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} $* * S t e p 5 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

= \frac{1}{\sqrt{0+4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
$1/4$ \lim_{x \to 5} \frac{x-5}{\sqrt{x-1} - 2} \cdot \frac{\sqrt{x-1} + 2}{\sqrt{x-1} + 2} $* * S t e p 3 : * * E x p a n d t h e d e n o m i n a t o r . >$

= \lim_{x \to 5} \frac{(x-5)(\sqrt{x-1} + 2)}{(x-1) - 4}
$>$

= \lim_{x \to 5} \frac{(x-5)(\sqrt{x-1} + 2)}{x-5}
$(since$ = \lim_{x \to 5} (\sqrt{x-1} + 2) $* * S t e p 5 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

= \sqrt{5-1} + 2 = \sqrt{4} + 2 = 2 + 2 = 4
$f(x)$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = (1)^2 = 1 $x \ge 1$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1 $\lim_{x \to 1^-} f(x) = 1$ \lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} (3-x) = 3 - 2 = 1 $x \to 2^+$ \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (x^2 - 1) = (2)^2 - 1 = 4 - 1 = 3 $\lim_{x \to 2^-} g(x) = 1$ \lim_{x \to \infty} \frac{\frac{3x^2}{x^2} - \frac{x}{x^2} + \frac{5}{x^2}}{\frac{2x^2}{x^2} + \frac{4x}{x^2} - \frac{1}{x^2}} $>$

= \lim_{x \to \infty} \frac{3 - \frac{1}{x} + \frac{5}{x^2}}{2 + \frac{4}{x} - \frac{1}{x^2}}
$\lim_{x \to \infty} \frac{k}{x^n} = 0$ = \frac{3 - 0 + 0}{2 + 0 - 0} $>$

= \frac{3}{2}
$3/2$ \lim_{x \to -\infty} \frac{\frac{4x^3}{x^4} - \frac{7x}{x^4}}{\frac{2x^4}{x^4} + \frac{3x^2}{x^4} - \frac{1}{x^4}} $>$

= \lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{7}{x^3}}{2 + \frac{3}{x^2} - \frac{1}{x^4}}
$\lim_{x \to \pm\infty} \frac{k}{x^n} = 0$ = \frac{0 - 0}{2 + 0 - 0} $>$

= \frac{0}{2}
$>$

= 0
$\lim_{x \to c} f(x) = \infty$ \lim_{x \to 3^+} \frac{x+1}{x-3} = \infty $\infty$ \lim_{x \to 0^-} \frac{x-2}{x^2} = -\infty $-\infty$ -x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2 $x \to 0$ \lim_{x \to 0} (-x^2) = 0 $>$

\lim_{x \to 0} (x^2) = 0
$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 $0$ \lim_{x \to 0} g(x) = \lim_{x \to 0} \left(1 - \frac{x^2}{2}\right) = 1 - \frac{0^2}{2} = 1 $>$

\lim_{x \to 0} h(x) = \lim_{x \to 0} (1) = 1
$x \to 0$ \lim_{x \to 0} \cos x = 1 $\lim_{x \to 0} \frac{\sin(4x)}{x}$ \lim_{x \to 0} \frac{\sin(4x)}{x} = \lim_{x \to 0} \frac{\sin(4x)}{x} \cdot \frac{4}{4} $>$

= \lim_{x \to 0} 4 \cdot \frac{\sin(4x)}{4x}
$. As$ = 4 \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} $* * S t e p 3 : * * A p p l y t h e s p e c i a l t r i g o n o m e t r i c l i m i t . >$

= 4 \cdot 1 = 4
$4$ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x}}{x} $>$

= \lim_{x \to 0} \frac{\sin x}{x \cos x}
$* * S t e p 2 : * * S e p a r a t e t h e l i m i t i n t o p r o d u c t s . >$

= \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{1}{\cos x} \right)
$>$

= \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{\cos x} \right)
$* * S t e p 3 : * * A p p l y t h e s p e c i a l t r i g o n o m e t r i c l i m i t a n d d i r e c t s u b s t i t u t i o n . >$

= (1) \cdot \left( \frac{1}{\cos 0} \right)
$>$

= 1 \cdot \frac{1}{1}
$>$

= 1
$e$ \lim_{x \to \infty} \left(1+\frac{2}{x}\right)^x = \lim_{x \to \infty} \left(1+\frac{1}{x/2}\right)^x $>$

= \lim_{x \to \infty} \left(1+\frac{1}{x/2}\right)^{(x/2) \cdot 2}
$>$

= \lim_{u \to \infty} \left(\left(1+\frac{1}{u}\right)^u\right)^2
$\lim_{x \to \infty} (f(x))^k = (\lim_{x \to \infty} f(x))^k$ = \left(\lim_{u \to \infty} \left(1+\frac{1}{u}\right)^u\right)^2 $e$ = (e)^2 = e^2 $e^2$ \lim_{x \to 0} (1+3x)^{1/x} = \lim_{u \to 0} (1+u)^{3/u} $* * S t e p 2 : * * U s e t h e p o w e r l a w f o r l i m i t s . >$

= \lim_{u \to 0} ((1+u)^{1/u})^3
$>$

= \left(\lim_{u \to 0} (1+u)^{1/u}\right)^3
$e$ = (e)^3 = e^3 $e^3$ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x \cos x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} $>$

= \lim_{x \to 0} \frac{(1+x) - 1}{x \cos x (\sqrt{1+x} + 1)}
$>$

= \lim_{x \to 0} \frac{x}{x \cos x (\sqrt{1+x} + 1)}
$x$ = \lim_{x \to 0} \frac{1}{\cos x (\sqrt{1+x} + 1)} $* * S t e p 4 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

= \frac{1}{\cos 0 (\sqrt{1+0} + 1)}
$>$

= \frac{1}{1 \cdot (1 + 1)}
$>$

= \frac{1}{2}
$1/2$ \lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{x \to 0} \frac{e^{2x} - e^0}{x - 0} $for$ f'(0) = 2e^{2 \cdot 0} = 2e^0 = 2 \cdot 1 = 2 $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$ \lim_{x \to 2} \frac{(x-2)(x+2)}{(x-1)(x-2)} $(since$ \lim_{x \to 2} \frac{x+2}{x-1} $* * S t e p 5 : * * A p p l y d i r e c t s u b s t i t u t i o n . >$

\frac{2+2}{2-1} = \frac{4}{1} = 4
$\lim_{x \to 0} \frac{\sin(5x)}{2x}$ \lim_{x \to 0} \frac{\sin(5x)}{2x} = \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin(5x)}{x} $* * S t e p 2 : * * I n t r o d u c e a f a c t o r o f 5 i n t h e d e n o m i n a t o r a n d c o m p e n s a t e i n t h e n u m e r a t o r . >$

= \lim_{x \to 0} \frac{1}{2} \cdot 5 \cdot \frac{\sin(5x)}{5x}
$>$

= \frac{5}{2} \lim_{x \to 0} \frac{\sin(5x)}{5x}
$. As$ = \frac{5}{2} \lim_{\theta \to 0} \frac{\sin \theta}{\theta} $* * S t e p 4 : * * A p p l y t h e s p e c i a l t r i g o n o m e t r i c l i m i t . >$

= \frac{5}{2} \cdot 1 = 2.5
$\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2x}}{3x - 1}$ \lim_{x \to -\infty} \frac{\frac{\sqrt{x^2 + 2x}}{x}}{\frac{3x - 1}{x}} $x \to -\infty$ \frac{\sqrt{x^2 + 2x}}{x} = \frac{\sqrt{x^2(1 + 2/x)}}{x} = \frac{\sqrt{x^2}\sqrt{1 + 2/x}}{x} = \frac{|x|\sqrt{1 + 2/x}}{x} $x \to -\infty$ = \frac{-x\sqrt{1 + 2/x}}{x} = -\sqrt{1 + 2/x} $And for the denominator:>$

\frac{3x - 1}{x} = 3 - \frac{1}{x}
$* * S t e p 3 : * * S u b s t i t u t e b a c k i n t o t h e l i m i t e x p r e s s i o n . >$

\lim_{x \to -\infty} \frac{-\sqrt{1 + 2/x}}{3 - 1/x}
$x \to -\infty$ = \frac{-\sqrt{1 + 0}}{3 - 0} = \frac{-\sqrt{1}}{3} = -\frac{1}{3} $-1/3$ f(x) = \begin{cases} x^2 & \text{if } x < 2 \\ 3x - 2 & \text{if } x \ge 2 \end{cases} $x \ge 2$ f(2) = 3(2) - 2 = 6 - 2 = 4 $\lim_{x \to 2} f(x)$ \lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{-}} x^2 = (2)^2 = 4 $,$ \lim_{x \to 2^{+}} f(x) = \lim_{x \to 2^{+}} (3x - 2) = 3(2) - 2 = 4 $\lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{+}} f(x) = 4$ f(x) = \begin{cases} ax + 3 & \text{if } x < 1 \\ x^2 + 2 & \text{if } x \ge 1 \end{cases} $x \ge 1$ f(1) = (1)^2 + 2 = 3 $x \to 1^{-}$ \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (ax + 3) = a(1) + 3 = a + 3 $x \to 1^{+}$ \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (x^2 + 2) = (1)^2 + 2 = 3 $, we need$ a + 3 = 3 $>$

a = 0
$g(x) = \frac{x^2 - 4}{x - 2}$ \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)
$>$

= 2+2 = 4
$h(x) = \begin{cases} x+1 & amp; \text{if } x & lt; 0 \\ x-1 & amp; \text{if } x \ge 0 \end{cases}$ h(0) = 0 - 1 = -1 $* * S t e p 2 : * * E v a l u a t e l e f t - h a n d a n d r i g h t - h a n d l i m i t s . L e f t - h a n d l i m i t : >$

\lim_{x \to 0^{-}} h(x) = \lim_{x \to 0^{-}} (x+1) = 0+1 = 1
$Right-hand limit:>$

\lim_{x \to 0^{+}} h(x) = \lim_{x \to 0^{+}} (x-1) = 0-1 = -1
$\lim_{x \to 0^{-}} h(x) \neq \lim_{x \to 0^{+}} h(x)$ \lim_{x \to 0} \frac{1}{x^2} $,$ \lim_{x \to 0} \frac{1}{x^2} = +\infty $x=2$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $is undefined, the discontinuity is removable. We could define$ x^2 - 4 = 0 $>$

(x-2)(x+2) = 0
$>$

x = 2 \quad \text{or} \quad x = -2
$and$ x-3 \ge 0 $>$

x \ge 3
$[3, \infty)$ \ln(x) = 0 $x = e^0 = 1$ \cos x \ge 0 $\cos x \ge 0$ b = \frac{\pi}{2} \approx \frac{3.14159}{2} \approx 1.570795 $b \approx 1.571$ f(0) = (0)^3 - 4(0) + 1 = 1 $>$

f(1) = (1)^3 - 4(1) + 1 = 1 - 4 + 1 = -2
$f(0) = 1$ f(x) = \begin{cases} x^2 - 1 & \text{if } x < 1 \\ ax + b & \text{if } 1 \le x < 3 \\ 5 & \text{if } x \ge 3 \end{cases} $x=1$ f(1) = a(1) + b = a + b $>$

\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (x^2 - 1) = (1)^2 - 1 = 0
$>$

\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (ax + b) = a(1) + b = a + b
$Setting these equal:>$

a + b = 0 \quad \text{(Equation 1)}
$x=3$ f(3) = 5 $>$

\lim_{x \to 3^{-}} f(x) = \lim_{x \to 3^{-}} (ax + b) = a(3) + b = 3a + b
$>$

\lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} (5) = 5
$Setting these equal:>$

3a + b = 5 \quad \text{(Equation 2)}
$and$ 3a + (-a) = 5 $>$

2a = 5
$>$

a = \frac{5}{2}
$a = \frac{5}{2}$ b = -\frac{5}{2} $to be continuous everywhere,$ f(x) = \begin{cases} \frac{\sin(3x)}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} $, the limit of$ \lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{3 \sin(3x)}{3x} $>$

= 3 \lim_{x \to 0} \frac{\sin(3x)}{3x}
$. As$ = 3 \lim_{y \to 0} \frac{\sin y}{y} $>$

= 3(1) = 3
$f(0)$ k = 3 $(0, \infty)$ x^2 > 1 $>$

|x| > 1
$or$ f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & \text{if } x \neq 1 \\ k & \text{if } x = 1 \end{cases} $, the limit of$ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x - 1} $>$

= \lim_{x \to 1} (x+1)
$>$

= 1+1 = 2
$f(1)$ k = 2 $f(x)$ f(1) = 5 $Left-hand limit:>$

\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x^2 = (1)^2 = 1
$Right-hand limit:>$

\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (2-x) = 2-1 = 1
$\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = 1$ x-3 = 0 $>$

x = 3
$' in math mode at position 4: At ̲x=3, the funct\dots" style="color:#cc0000">At x=3$, the function is undefined, hence discontinuous. Step 3: Determine the type of discontinuity (optional for this question, but good practice). >$ \lim_{x \to 3} \frac{x^2 - 9}{x-3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} $>$

= \lim_{x \to 3} (x+3)
$>$

= 3+3 = 6
\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{x^3} $" options=["$ -1/3 $", "$ 0 $", "$ 1/3 $", "Does not exist"] answer="$ -1/3 $" hint="This is an indeterminate form of type$ 0/0 $. Apply L'Hôpital's Rule repeatedly." solution="Applying L'Hôpital's Rule:$ \lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{x^3} \quad \left(\frac{0}{0}\right) $First application:$ \lim_{x \to 0} \frac{(\cos(x) - x \sin(x)) - \cos(x)}{3x^2} = \lim_{x \to 0} \frac{-x \sin(x)}{3x^2} = \lim_{x \to 0} \frac{-\sin(x)}{3x} \quad \left(\frac{0}{0}\right) $Second application:$ \lim_{x \to 0} \frac{-\cos(x)}{3} = \frac{-\cos(0)}{3} = \frac{-1}{3} $"::: :::question type="NAT" question="For what value of$ k $is the function$ f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases} $continuous at$ x=3 $?" answer="6" hint="For continuity at$ x=3 $, the limit of$ f(x) $as$ x \to 3 $must equal$ f(3) $. Simplify the rational expression." solution="For$ f(x) $to be continuous at$ x=3 $, we must have$ \lim_{x \to 3} f(x) = f(3) $.First, find the limit:$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) $Substituting$ x=3 $into the simplified expression:$ \lim_{x \to 3} (x+3) = 3+3=6 $For continuity,$ f(3) $must equal this limit. Since$ f(3)=k $, we have$ k=6 $.Thus,$ k=6 $."::: :::question type="MCQ" question="Determine the horizontal asymptotes of the function$ f(x) = \frac{\sqrt{9x^2+2}}{x-1} $." options=["$ y=3 $only", "$ y=-3 $only", "$ y=3 $and$ y=-3 $' in math mode at position 40: \dotsotes & quot;] answer= & quot; ̲y=3" style="color:#cc0000">", "No horizontal asymptotes"] answer="$y=3$ and $y=-3$ " hint="Evaluate $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ . Remember that $\sqrt{x^2}=|x|$ , which is $x$ for $x>0$ and $-x$ for $x<0$ ." solution="To find horizontal asymptotes, we evaluate the limits as $x \to \infty$ and $x \to -\infty$ .

For
$\lim_{x \to \infty} f(x)$ :
$\lim_{x \to \infty} \frac{\sqrt{9x^2+2}}{x-1} = \lim_{x \to \infty} \frac{\sqrt{x^2(9+2/x^2)}}{x(1-1/x)} = \lim_{x \to \infty} \frac{|x|\sqrt{9+2/x^2}}{x(1-1/x)}$
Since $x \to \infty$ , $x > 0$ , so $|x|=x$ :
$\lim_{x \to \infty} \frac{x\sqrt{9+2/x^2}}{x(1-1/x)} = \lim_{x \to \infty} \frac{\sqrt{9+2/x^2}}{1-1/x} = \frac{\sqrt{9+0}}{1-0} = \frac{3}{1} = 3$
So, $y=3$ is a horizontal asymptote.

For
$\lim_{x \to -\infty} f(x)$ :
$\lim_{x \to -\infty} \frac{\sqrt{9x^2+2}}{x-1} = \lim_{x \to -\infty} \frac{|x|\sqrt{9+2/x^2}}{x(1-1/x)}$
Since $x \to -\infty$ , $x < 0$ , so $|x|=-x$ :
$\lim_{x \to -\infty} \frac{-x\sqrt{9+2/x^2}}{x(1-1/x)} = \lim_{x \to -\infty} \frac{-\sqrt{9+2/x^2}}{1-1/x} = \frac{-\sqrt{9+0}}{1-0} = \frac{-3}{1} = -3$
So, $y=-3$ is also a horizontal asymptote.

Therefore, the horizontal asymptotes are
$y=3$ and $y=-3$ ."
:::

:::question type="NAT" question="Let
$f(x) = x^3 - 4x + 1$ . How many distinct roots are guaranteed to exist for $f(x)$ in the interval $[-2, 2]$ by the Intermediate Value Theorem?" answer="2" hint="Evaluate $f(x)$ at the endpoints and at any points within the interval where the function might change sign. Look for sign changes." solution="We evaluate the function at the endpoints of the interval and at strategic points to identify sign changes.
$f(x) = x^3 - 4x + 1$
$f(-2) = (-2)^3 - 4(-2) + 1 = -8 + 8 + 1 = 1$
$f(1) = (1)^3 - 4(1) + 1 = 1 - 4 + 1 = -2$
$f(2) = (2)^3 - 4(2) + 1 = 8 - 8 + 1 = 1$

Since $f(-2) = 1 > 0$ and $f(1) = -2 < 0$ , and $f(x)$ is a polynomial (hence continuous everywhere), by the Intermediate Value Theorem, there must be at least one root in the interval $(-2, 1)$ .

Since $f(1) = -2 < 0$ and $f(2) = 1 > 0$ , by the Intermediate Value Theorem, there must be at least one root in the interval $(1, 2)$ .

These two intervals are disjoint, so they guarantee two distinct roots in
$[-2, 2]$ .
To confirm there are no more roots guaranteed solely by IVT with these specific points, we can consider the intervals. However, the question asks for the minimum number of roots guaranteed. We have found two such roots.
Thus, 2 distinct roots are guaranteed."
:::

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What's Next?

💡 Continue Your CMI Journey

Having established a solid understanding of limits and continuity, you are now equipped with the fundamental tools necessary for the next major pillars of Calculus. The concepts of limits are directly applied in the definition of the derivative, which will be the focus of the subsequent chapters. Continuity is a prerequisite for many theorems in differentiation and integration, ensuring the well-behaved nature of functions we analyze. Prepare to explore rates of change, optimization, and the foundational calculus of integrals.

Limits and Continuity

Limits and Continuity

Chapter Contents

| Topic |

Part 1: Limits

Core Concepts

1. Intuitive Definition of a Limit

2. Formal Definition of a Limit (Epsilon-Delta)

3. Limit Laws

4. Direct Substitution Property

5. Factoring and Cancellation

6. Rationalization

7. One-Sided Limits

8. Limits at Infinity

9. Infinite Limits

10. Squeeze Theorem (Sandwich Theorem)

11. Special Trigonometric Limits

12. Limits involving the number eee

Advanced Applications

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Part 2: Continuity

Core Concepts

1. Definition of Continuity

2. Types of Discontinuity

3. Continuity of Common Functions

4. Continuity of Composite Functions

5. Intermediate Value Theorem (IVT)

6. Extreme Value Theorem (EVT)

Advanced Applications

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Calculus

Integrals

Techniques of Integration

Applications of Integrals

Partial Derivatives

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Study Notes

Short Notes

Test Series

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Previous Year Papers

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12. Limits involving the number $e$