Exercise 12.1 (Revised) - Chapter 13 - Limits & Derivatives - Ncert Solutions class 11 - Maths
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Chapter 12: Limits & Derivatives - NCERT Solutions for Class 11 Maths
Evaluate the following limits in Exercises 1 to 22.
Ex 12.1 Question 1.
$\lim _{x \rightarrow 3} x+3$
Answer.
$\lim _{x \rightarrow 3} x+3=3+3=6$
Ex 12.1 Question 2.
$\lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right)$
Answer.
$\lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right)=\left(\pi-\frac{22}{7}\right)$
Ex 12.1 Question 3.
$\lim _{r \rightarrow 1} \pi r^2$
Answer.
$\lim _{r \rightarrow 1} \pi r^2=\pi \times(1)^2=\pi$
Ex 12.1 Question 4.
$\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}$
Answer.
$\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4 x 4+3}{4-2}=\frac{19}{2}$
Ex 12.1 Question 5.
$\lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1}$
Answer:
$\text {} \lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1}=\frac{(-1)^{10}+(-1)^5+1}{-1-1}=\frac{1-1+1}{-2}=\frac{-1}{2}$
Ex 12.1 Question 6.
$\lim _{x \rightarrow 0} \frac{(x+1)^5-1}{x}$
Answer.
$\lim _{x \rightarrow 0} \frac{(x+1)^5-1}{x}$ is of the form $\frac{0}{0}$
Put $x+1=y$, now as $x \rightarrow 0, y \rightarrow 1$
$\therefore \lim _{x \rightarrow 0}\left(\frac{(x+1)^5-1}{x}\right)=\lim _{y \rightarrow 1}\left(\frac{y^5-1}{y-1}\right)=\lim _{y \rightarrow 1}\left(\frac{y^5-1^5}{y-1}\right)$
$=5 \cdot 1^{5-1}=5 \cdot 1=5 \quad$ since $\lim _{x \rightarrow a}\left(\frac{x^n-a^n}{x-a}\right)=n a^{n-1}$
$\therefore \lim _{x \rightarrow 0} \frac{(x+1)^5-1}{x}=5$
Ex 12.1 Question 7.
$\lim _{x \rightarrow 2} \frac{3 x^2-x-10}{x^2-4}$
Answer.
$\lim _{x \rightarrow 2} \frac{3 x^2-x-10}{x^2-4}\left[\frac{0}{0}\right.$ form $]$
$=\lim _{x \rightarrow 2} \frac{(x-2)(3 x+5)}{(x+2)(x-2)}$
$
=\lim _{x \rightarrow 2} \frac{(3 x+5)}{(x+2)}=\frac{6+5}{2+2}=\frac{11}{4}
$
Ex 12.1 Question 8.
$\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}$
Answer.
$\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}$ is of the form $\frac{0}{0}$
$
\begin{aligned}
& \therefore \lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}=\lim _{x \rightarrow 3} \frac{\left(x^2+9\right)(x+3)(x-3)}{(x-3)(2 x+1)} \\
& =\lim _{x \rightarrow 3} \frac{\left(x^2+9\right)(x+3)}{(2 x+1)} \\
& =\frac{(9+9)(3+3)}{(2 \times 3+1)}=\frac{18 \times 6}{7}=\frac{108}{7}
\end{aligned}
$
Ex 12.1 Question 9.
$\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}$
Answer.
$\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a \times 0+b}{c \times 0+1}=b$
Ex 12.1 Question 10.
$\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$
Answer.
$\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$ is of the form $\frac{0}{0}$
$\begin{aligned}
& \therefore \lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim _{z \rightarrow 1} \frac{\left(z^{\frac{1}{6}}\right)^2-(1)^2}{z^{\frac{1}{6}}-1} \\
& =\lim _{z \rightarrow 1} \frac{\left(z^{\frac{1}{6}}+1\right)\left(z^{\frac{1}{6}}-1\right)}{z^{\frac{1}{6}}-1}
\end{aligned}$
$
\begin{aligned}
& =\lim _{z \rightarrow 1}\left(z^{\frac{1}{6}}+1\right) \\
& =(1)^{\frac{1}{6}}+1=1+1=2
\end{aligned}
$
Ex 12.1 Question 11.
$\lim _{x \rightarrow 1} \frac{a x^2+b x+c}{c x^2+b x+a}, a+b+c \neq 0$
Answer.
$\lim _{x \rightarrow 1} \frac{a x^2+b x+c}{c x^2+b x+a}$
$
\begin{aligned}
& =\frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} \\
& =\frac{a+b+c}{c+b+a}=1
\end{aligned}
$
Ex 12.1 Question 12.
$\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$
Answer.
$\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}=\lim _{x \rightarrow-2} \frac{\frac{x+2}{2 x}}{x+2}$
$
\begin{aligned}
& =\lim _{x \rightarrow-2} \frac{x+2}{2 x} \times \frac{1}{x+2} \\
& =\lim _{x \rightarrow-2} \frac{1}{2 x}=\frac{1}{2 \times(-2)}=\frac{-1}{4}
\end{aligned}
$
Ex 12.1 Question 13.
$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}$
$
\begin{aligned}
& \text { Ans. } \lim _{x \rightarrow 0} \frac{\sin a x}{b x} \\
& =\lim _{x \rightarrow 0}\left(\frac{\sin a x}{a x} \times \frac{a x}{b x}\right) \\
& =\frac{a}{b} \lim _{x \rightarrow 0} \frac{\sin a x}{a x} \\
& =\frac{a}{b} \lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right) \quad[x \rightarrow 0 \Rightarrow a x \rightarrow 0] \text { and }\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\
& =\frac{a}{b} \times 1=\frac{a}{b}
\end{aligned}
$
Ex 12.1 Question 14.
$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$
Answer.
$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}$
$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\frac{\sin a x}{a x}\right) a x}{\left(\frac{\sin b x}{b x}\right) b x}=\frac{a}{b} \lim _{x \rightarrow 0} \frac{\left(\frac{\sin a x}{a x}\right)}{\left(\frac{\sin b x}{b x}\right)} \\
& =\frac{a}{b} \cdot \frac{\lim _{x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)}{\lim _{b x \rightarrow 0}\left(\frac{\sin b x}{b x}\right)} \text { since }\left[\begin{array}{l}
x \rightarrow 0 \Rightarrow a x \rightarrow 0 \\
x \rightarrow 0 \Rightarrow b x \rightarrow 0
\end{array}\right]
\end{aligned}
$
$
=\frac{a}{b} \cdot \frac{1}{1}=\frac{a}{b} \quad\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]
$
Ex 12.1 Question 15.
$\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$
Answer.
$\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\left[\frac{0}{0}\right.$ form $]$
Put $x=\pi+y$ now as $x \rightarrow \pi=y \rightarrow 0$
$
\begin{aligned}
& \therefore \lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}=\lim _{y \rightarrow 0} \frac{\sin (\pi-\pi-y)}{\pi(\pi-\pi-y)} \\
& =\lim _{y \rightarrow 0} \frac{\sin (-y)}{-\pi y}=\lim _{y \rightarrow 0} \frac{-\sin y}{-\pi y} \quad[\sin (-\theta)=-\sin \theta] \\
& =\frac{1}{\pi} \lim _{y \rightarrow 0} \frac{\sin y}{y} \quad\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\
& =\frac{1}{\pi} \times 1=\frac{1}{\pi}
\end{aligned}
$
Ex 12.1 Question 16.
$\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}$
Answer.
$\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}$
Ex 12.1 Question 17.
$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$
Answer.
$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$ is of the form $\frac{0}{0}$
$
\therefore \lim _{x \rightarrow 0}\left(\frac{\cos 2 x-1}{\cos x-1}\right)=\lim _{x \rightarrow 0} \frac{\left(2 \cos ^2 x-1\right)-1}{\cos x-1} \quad\left[\cos 2 \theta=2 \cos ^2 \theta-1\right]
$
$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{2\left(\cos ^2 x-1\right)}{\cos x-1} \\
& =2 \lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x+1)}{(\cos x-1)} \\
& =2 \lim _{x \rightarrow 0}(\cos x+1)=2(1+1)=2 \times 2=4
\end{aligned}
$
Ex 12.1 Question 18.
$\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}$
Answer:
$
\begin{aligned}
& \text {} \lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}\left[\frac{0}{0} \text { form }\right] \\
& =\frac{1}{b} \lim _{x \rightarrow 0} \frac{x(a+\cos x)}{\sin x} \\
& =\frac{1}{b} \lim _{x \rightarrow 0} \frac{x}{\sin x} \cdot \lim _{x \rightarrow 0}(a+\cos x) \\
& =\frac{1}{b} \cdot \frac{\lim _{x \rightarrow 0}(a+\cos x)}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)} \quad\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\
& =\frac{1}{b} \times \frac{a+1}{1}=\frac{a+1}{b}
\end{aligned}
$
Ex 12.1 Question 19.
$\lim _{x \rightarrow 0} x \sec x$
Answer.
$\lim _{x \rightarrow 0} x \sec x=\lim _{x \rightarrow 0} x \frac{1}{\cos x}$
$
=\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{1}=0
$
Ex 12.1 Question 20.
$\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}, a=b, a+b \neq 0$
Answer.
$\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}\left[\frac{0}{0}\right.$ form $]$
Dividing numerator and denominator by $\alpha x$ :
$
=\lim _{x \rightarrow 0} \frac{\frac{\sin a x}{a x}+\frac{b x}{a x}}{\frac{a x}{a x}+\frac{\sin b x}{a x}}
$
$\begin{array}{ll}
=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin a x}{a x}\right)+\frac{b x}{a x}}{1+\left(\frac{\sin b x}{b x}\right) \frac{b x}{a x}} & \\
=\frac{\lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)+\frac{b}{a} \lim _{a x \rightarrow 0} 1}{1+\lim _{b x \rightarrow 0}\left(\frac{\sin b x}{b x}\right) \cdot \frac{b}{a} \lim _{b x \rightarrow 0} 1} & {\left[\begin{array}{l}
x \rightarrow 0 \Rightarrow a x \rightarrow 0 \\
x \rightarrow 0 \Rightarrow b x \rightarrow 0
\end{array}\right]} \\
=\frac{1+\frac{b}{a}}{1+\frac{b}{a}}=1 & {\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]}
\end{array}$
Ex 12.1 Question 21.
$\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)$
Answer.
Given: $\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)$
$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)=\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \\
& =\lim _{x \rightarrow 0} \tan \frac{x}{2}=0
\end{aligned}
$
Ex 12.1 Question 22.
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$
Answer.
Given: $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\left[\frac{0}{0}\right.$ form $]$
$\text { Put } x=\frac{\pi}{2}+y \text { now as } x \rightarrow \frac{\pi}{2}, y \rightarrow 0$
$\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\tan 2\left(\frac{\pi}{2}+y\right)}{\frac{\pi}{2}+y-\frac{\pi}{2}} \\
& =\lim _{y \rightarrow 0} \frac{\tan (\pi+2 y)}{y}=\lim _{y \rightarrow 0} \frac{\tan 2 y}{y} \quad[\tan (\pi+\theta)=\tan \theta] \\
& =\lim _{y \rightarrow 0} \frac{\tan 2 y}{2 y} \times 2 \\
& =2 \lim _{y \rightarrow 0} \frac{\tan 2 y}{2 y} \quad\left[\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\right] \\
& =2 \times 1=2 \\
&
\end{aligned}$
Ex 12.1 Question 23.
Find $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$ where $f(x)= \begin{cases}2 x+3 & x \leq 0 \\ 3(x+1) & x>0\end{cases}$
Answer.
Given: $f(x)= \begin{cases}2 x+3 & x \leq 0 \\ 3(x+1) & x>0\end{cases}$
For $\mathrm{x}>0$ Right hand limit $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 3(x+1)=3(0+1)=3$
For $\mathrm{x}<0 \quad$ Left hand limit $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(2 x+3)=2(0)+3=3$
As $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=3$, we have $\lim _{x \rightarrow 0} f(x)=3$
For $\mathrm{x}>1$ Right hand limit $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 3(x+1)=3(1+1)=6$
For $\mathrm{x}<1 \quad$ Left hand limit $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 3(x+1)=3(1+1)=6$
As $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=6$, we have $\lim _{x \rightarrow 1} f(x)=6$
Ex 12.1 Question 24.
Find $\lim _{x \rightarrow 1} f(x)$ where $f(x)= \begin{cases}x^2-1 & x \leq 1 \\ -x^2-1 & x>1\end{cases}$
Answer.
Given: $f(x)= \begin{cases}x^2-1 & x \leq 1 \\ -x^2-1 & x>1\end{cases}$
For $\mathrm{x}>1 \quad$ Right hand limit $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(-x^2-1\right)=-1-1=-2$
For $\mathrm{x}<1 \quad$ Left hand limit $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^2-1\right)=1-1=0$
As $\lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x)$,we have $\lim _{x \rightarrow 1} f(x)$ does not exist
Ex 12.1 Question 25.
Evaluate $\lim _{x \rightarrow 0} f(x)$ where $f(x)= \begin{cases}\frac{|x|}{x} & x \neq 0 \\ 0 & x=0\end{cases}$
Answer.
Given: $f(x)= \begin{cases}\frac{|x|}{x} & x \neq 0 \\ 0 & x=0\end{cases}$
We have $|x|=x$ when $\mathrm{x}$ is positive
$\therefore$ For $\mathrm{x}>0$ Right hand limit $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\lim _{x \rightarrow 0} \frac{x}{x}=1$
$\text { As } \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \text {,we have } \lim _{x \rightarrow 0} f(x) \text { does not exist }$
Ex 12.1 Question 26.
Find $\lim _{x \rightarrow 0} f(x)$ where $f(x)= \begin{cases}\frac{x}{|x|} & x \neq 0 \\ 0 & x=0\end{cases}$
Answer.
Given: $f(x)= \begin{cases}\frac{x}{|x|} & x \neq 0 \\ 0 & x=0\end{cases}$
We have $|x|=x$ when $\mathrm{x}$ is positive
$\therefore$ For $\mathrm{x}>0$ Right hand limit $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{x}{|x|}=\lim _{x \rightarrow 0} \frac{x}{x}=1$
We have $|x|=-x$ when $\mathrm{x}$ is negative
$\therefore$ For $\mathrm{x}<0 \quad$ Left hand limit $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|}=\lim _{x \rightarrow 0} \frac{x}{-x}=-1$
As $\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$,we have $\lim _{x \rightarrow 0} f(x)$ does not exist
Ex 12.1 Question 27.
Find $\lim _{x \rightarrow 5} f(x)$ where $f(x)=|x|-5$
Answer.
Given: $f(x)=|x|-5$
L.H.L. $\lim _{x \rightarrow 5^{}} f(x)=\lim _{x \rightarrow 5}|x|-5$
Putting $x=5-h$ as $x \rightarrow 5, h \rightarrow 0$
$
\begin{aligned}
& \therefore \lim _{h \rightarrow 0}|5-h|-5=\lim _{h \rightarrow 0} 5-h-5 \\
& =\lim _{h \rightarrow 0}(-h)=0
\end{aligned}
$
R.H.L. $\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{+}}|x|-5$
Putting $x=5+h$ as $x \rightarrow 5, h \rightarrow 0$
$
\begin{aligned}
& \therefore \lim _{h \rightarrow 0}|5+h|-5=\lim _{h \rightarrow 0} 5+h-5 \\
& =\lim _{h \rightarrow 0} h=0
\end{aligned}
$
Here, L.H.L. = R.H.L.
Therefore, this limit exists at $x=5$ and $\lim _{x \rightarrow 5} f(x)=0$
Ex 12.1 Question 28.
Suppose $f(x)=\left\{\begin{array}{cc}a+b x & x<1 \\ 4 & x=1 \\ b-a x & x>1\end{array}\right.$ and if $\lim _{x \rightarrow 1} f(x)=f(1)$ what are possible values of $a$ and $b$ ?
Answer.
Given: $f(x)=\left\{\begin{array}{cl}a+b x & x<1 \\ 4 & x=1 \\ b-a x & x>1\end{array}\right.$ and $\lim _{x \rightarrow 1} f(x)=f(1)$
$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)=4 \\
& \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=4 \text { and } \lim _{x \rightarrow 1^{-}} f(x)=4 \\
& \Rightarrow \lim _{x \rightarrow 1} a+b x=4 \text { and } \lim _{x \rightarrow 1} b-a x=4 \\
& \Rightarrow a+b=4 \text { and } b-a=4
\end{aligned}
$
On solving these equation, we get $a=0$ and $b=4$
Ex 12.1 Question 29.
Let $a_1, a_2, \ldots \ldots . a_n$ be fixed real numbers and define a function $f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$. What is $\lim _{x \rightarrow a_1} f(x)$ ? For some $a \neq a_1, a_2, \ldots \ldots a_n$ : compute $\lim _{x \rightarrow a} f(x)$.
Answer:
$\text {Given: } f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$
$
\begin{aligned}
& \text { Now } \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right) \\
& =\left(a_1-a_1\right)\left(a_1-a_2\right) \ldots\left(a_1-a_n\right) \\
& =0 \times\left(a_1-a_2\right) \ldots\left(a_1-a_n\right)=0 \\
& \therefore \lim _{x \rightarrow a} f(x)=0
\end{aligned}
$
Also $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$
$
\begin{aligned}
& =\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right) \\
& \therefore \lim _{x \rightarrow a} f(x)=\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right)
\end{aligned}
$
Ex 12.1 Question 30.
If $f(x)=\left\{\begin{array}{cl}|x|+1 & x<0 \\ 0 & x=0 \\ |x|-1 & x>0\end{array}\right.$ for what values of $a$ does $\lim _{x \rightarrow a} f(x)$ exists?
Answer.
Given: $f(x)=\left\{\begin{array}{cl}|x|+1 & x<0 \\ 0 & x=0 \\ |x|-1 & x>0\end{array} \Rightarrow f(x)=\left\{\begin{array}{cc}-x+1 & x<0 \\ 0 & x=0 \\ x-1 & x>0\end{array}\right.\right.$
Consider $\lim _{x \rightarrow a} f(x)$
When $a=0$
L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}|x|+1=\lim _{x \rightarrow 0}(-x+1)=(0+1)=1$
Also R.H.L. $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}|x|-1=\lim _{x \rightarrow 0}(x-1)=(0-1)=-1$
Here, L.H.L. $\neq$ R.H.L.
Therefore, this limit does not exist at $a=0$
When $\mathrm{a}>0$,
L.H.L. $=\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a}(x-1)=a-1 \quad\left[x \rightarrow a^{-} \Rightarrow 0
Also R.H.L. $=\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a}(x-1)=a-1 \quad\left[x \rightarrow a^{+} \Rightarrow 0 Here, L.H.L. = R.H.L.
Therefore, this limit exist at $\mathrm{x}=\mathrm{a}$ when $\mathrm{a}>0$
When $\mathrm{a}>0$,
L.H.L. $=\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a}(-x+1)=-a+1 \quad\left[x \rightarrow a^{-} \Rightarrow x
Also R.H.L. $=\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a}(-x+1)=-a+1 \quad\left[x \rightarrow a^{+} \Rightarrow a Here, L.H.L. $=$ R.H.L.
Therefore, this limit exist at $\mathrm{x}=\mathrm{a}$ when $\mathrm{a}<0$
$\therefore \lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$
Ex 12.1 Question 31.
If the function $f(x)$ satisfies $\lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$, evaluate $\lim _{x \rightarrow 1} f(x)$.
Answer.
$\lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$
$
\begin{aligned}
\Rightarrow & \frac{\lim _{x \rightarrow 1} f(x)-2}{\lim _{x \rightarrow 1}\left(x^2-1\right)}=\pi \\
\Rightarrow & \lim _{x \rightarrow 1} f(x)-2=\pi \cdot \lim _{x \rightarrow 1}\left(x^2-1\right)
\end{aligned}
$
Since $\lim _{x \rightarrow 1}\left(x^2-1\right)=(1)^2-1=1-1=0$, we get
$
\lim _{x \rightarrow 1} f(x)-2=\pi \cdot 0=0
$
$
\Rightarrow \lim _{x \rightarrow 1} f(x)=2
$
Ex 12.1 Question 32.
If $f(x)=\left\{\begin{array}{lc}m x^2+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^3+m, & x>1\end{array}\right.$ for what integer $m$ and $n$ does both $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$ exist?
Answer.
Left hand limit $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(m x^2+n\right)=m \cdot 0+n=n$
Right hand limit $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(n x+m)=n \cdot 0+m=m$
Thus $\lim _{x \rightarrow 0} f(x)$ exists only if $\mathrm{m}=\mathrm{n}$
Left hand limit $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(n x+m)=n \cdot 1+m=n+m$
Right hand limit $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(n x^3+m\right)=n \cdot 1+m=n+m$
As $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=m+n$ we get $\lim _{x \rightarrow 1} f(x)$ exist for any integral value of $m$ and n
