Exercise 9.4 - Chapter 9 Limits and Continuity 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 9.4
Evaluate the following limits
Question 1.
$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{7 x}
$
Solution:
$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{7 x}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{x}\right)^x\right]^7=e^7
$
Question 2.
$
\lim _{x \rightarrow 0}(1+x)^{1 / 3 x}
$
Solution:
$
\lim _{x \rightarrow 0}(1+x)^{\frac{1}{3 x}}=\lim _{x \rightarrow 0}\left[(1+x)^{\frac{1}{x}}\right]^{\frac{1}{3}}=e^{1 / 3}
$
Question 3.
$
\lim _{x \rightarrow \infty}\left(1+\frac{k}{x}\right)^{\frac{m}{x}}
$
Solution:
$
\lim _{x \rightarrow \infty}\left(1+\frac{k}{x}\right)^{\frac{m}{x}}=\lim _{x \rightarrow \infty}\left(1+\frac{k}{x}\right)^{\frac{m}{x}}=(1+0)=1
$
Question 4.
$
\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3}{2 x^2+5}\right)^{8 x^2+3}
$

Solution:
$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{2 x^2+3}{2 x^2+5}\right)^{8 x^2+3}=\lim _{x \rightarrow \infty}\left(1+\frac{(-2)}{2 x^2+5}\right)^{8 x^2+3+17-17} \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{-2}{2 x^2+5}\right)^{8 x^2+20} \times\left(1+\frac{-2}{2 x^2+5}\right)^{-17} \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{-2}{2 x^2+5}\right)^{4\left(2 x^2+5\right)} \times \lim _{x \rightarrow \infty}\left(1+\frac{-2}{2 x^2+5}\right)^{-17}=\lim _{x \rightarrow \infty}\left(1+\frac{-2}{y}\right)^{4 y} \lim _{x \rightarrow \infty}\left(1+\frac{-2}{y}\right)^{-17} \\
& =\left(e^{-2}\right)^4(1)=e^{-8}=\frac{1}{e^8}
\end{aligned}
$
Question 5.
$
\lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^{x+2}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^{x+2} & =\lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^x \cdot\left(1+\frac{3}{x}\right)^2 \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^x \lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^2 \\
& =e^3 \cdot(1)^2=e^3
\end{aligned}
$

Question 6.
$
\lim _{x \rightarrow 0} \frac{\sin ^3\left(\frac{x}{2}\right)}{x^3}
$
Solution:
$
\lim _{x \rightarrow 0} \frac{\sin ^3\left(\frac{x}{2}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{\sin ^3(x / 2)}{\frac{x}{8} \times 8}=\frac{1}{8} \lim _{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)=\frac{1}{8}
$
Question 7.
$
\lim _{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x}
$

Solution:
$
\lim _{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x}=\lim _{x \rightarrow 0} \frac{\frac{\sin \alpha x}{\alpha x} \times \alpha x}{\frac{\sin \beta x}{\beta x} \times \beta x}=\frac{\alpha}{\beta} \frac{\lim _{x \rightarrow 0}\left(\frac{\sin \alpha x}{\alpha x}\right)}{\lim _{x \rightarrow 0} \frac{\sin \beta x}{\beta x}}=\frac{\alpha}{\beta}
$
Question 8 .
$
\lim _{x \rightarrow 0} \frac{\tan 2 x}{\sin 5 x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\tan 2 x}{\sin 5 x} & =\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x}{2 x} \times 2 x}{\frac{\sin 5 x}{5 x} \times 5 x}=\frac{2}{5} \frac{\lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x}}{\lim _{x \rightarrow 0} \frac{\sin 5 x}{(5 x)}} \\
& =2 / 5
\end{aligned}
$

Question 9.
$
\lim _{\alpha \rightarrow 0} \frac{\sin \left(\alpha^n\right)}{(\sin \alpha)^m}
$
Solution:
$
\begin{aligned}
& \lim _{\alpha \rightarrow 0} \frac{\sin \left(\alpha^n\right)}{(\sin \alpha)^m}=\frac{\lim _{\alpha \rightarrow 0} \frac{\sin \left(\alpha^n\right)}{\alpha^n} \times \alpha^n}{\lim _{\alpha \rightarrow 0}\left(\frac{\sin \alpha}{\alpha} \times \alpha\right)^m} \\
& =\frac{\lim _{\alpha \rightarrow 0} \frac{\sin \left(\alpha^n\right)}{\alpha^n} \times \alpha^n}{\lim _{\alpha \rightarrow 0}\left(\frac{\sin \alpha}{\alpha}\right)^m \alpha^m} \\
& =\alpha^{n-m}(1)=\alpha \\
& \lim _{\alpha \rightarrow 0} \frac{\sin \left(\alpha^n\right)}{(\sin \alpha)^m}=\lim _{\alpha \rightarrow 0} \frac{\alpha^n}{\alpha^m}=\left\{\begin{array}{cll}
1 & \text { if } & m=n \\
0 & \text { if } & m>n \\
f(\alpha) \rightarrow \infty \text { as } \alpha \rightarrow 0 \text { if } m \end{array}\right. \\
&
\end{aligned}
$
Question 10 .
$
\lim _{x \rightarrow 0} \frac{\sin (a+x)-\sin (a-x)}{x}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin (a+x)-\sin (a-x)}{x} & =\frac{2 \cos \left(\frac{a+x+a-x}{2}\right) \sin \left(\frac{a+x-a+x}{2}\right)}{x} \\
& =\lim _{x \rightarrow 0} \frac{2 \cos (a) \sin (x)}{x} \\
& =2 \cos a \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)=2 \cos a
\end{aligned}
$
Question 11.
$
\lim _{x \rightarrow 0} \frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b} \times \frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+a^2}+a} \times \frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+b^2}+b} \\
&=\lim _{x \rightarrow 0} \frac{\left(x^2+a^2-a^2\right) \sqrt{x^2+b^2}+b}{\left(x^2+b^2-b^2\right) \sqrt{x^2+a^2}+a} \\
&=\lim _{x \rightarrow 0} \frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}=\frac{b+b}{a+a}=\frac{2 b}{2 a}=b / a
\end{aligned}
$
Question 12.
$
\lim _{x \rightarrow 0} \frac{2 \arcsin x}{3 x}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{2 \arcsin x}{3 x} & =\lim _{x \rightarrow 0} \frac{2 \sin ^{-1} x}{3 x} \\
& =\lim _{x \rightarrow 0} \frac{2 y}{3 \sin y}
\end{aligned}
$
Let $y=\sin ^{-1} x$
Let $y=\sin ^{-1} x$, as $x \rightarrow 0 y \rightarrow 0$
$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{2}{3\left(\frac{\sin y}{y}\right)} \\
& =\frac{2}{3} \lim _{y \rightarrow 0} \frac{1}{\left(\frac{\sin y}{y}\right)}=2 / 3
\end{aligned}
$
Question 13.
$
\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} & =\lim _{x \rightarrow 0} \frac{2 \sin ^2 x / 2}{x^2}=\lim _{x \rightarrow 0} \frac{\sin ^2 x / 2}{x^2 / 2} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 x / 2}{\frac{x^2}{4} \times 2}=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)^2 \\
& =\frac{1}{2}(1)=1 / 2
\end{aligned}
$

Question 14 .
$
\lim _{x \rightarrow 0} \frac{\tan 2 x}{x}
$
Solution:
$
\lim _{x \rightarrow 0} \frac{\tan 2 x}{x}=\lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \times 2=2 \lim _{x \rightarrow 0}\left(\frac{\tan 2 x}{2 x}\right)=2
$

Question 15 .
$
\lim _{x \rightarrow 0} \frac{2^x-3^x}{x}
$
Solution:
$
\lim _{x \rightarrow 0} \frac{2^x-3^x}{x}=\lim _{x \rightarrow 0} \frac{2^x}{x}-\lim _{x \rightarrow 0} \frac{3^x}{x}=\log 2-\log 3=\log \frac{2}{3}
$
Question 16.
$
\lim _{x \rightarrow 0} \frac{3^x-1}{\sqrt{x+1}-1}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{3^x-1}{\sqrt{x+1}-1} \times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} & =\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)(\sqrt{x+1}+1)}{x+1-1} \\
& =\lim _{x \rightarrow 0} \frac{3^x-1}{x} \cdot \lim _{x \rightarrow 0} \sqrt{x+1}+1 \\
& =(\log 3) 2 \\
& =2 \log 3=\log 3^2=\log 9
\end{aligned}
$
Question 17.
$
\lim _{x \rightarrow 0} \frac{1-\cos ^2 x}{x \sin 2 x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{1-\cos ^2 x}{x \sin 2 x}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x 2 \sin x \cos x} & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x \cos x} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0}\left(\frac{1}{2 \cos x}\right) \\
& =(1) \times \frac{1}{2}=1 / 2
\end{aligned}
$

Question 18.
$
\lim _{x \rightarrow \infty} x\left[3^{\frac{1}{x}}+1-\cos \left(\frac{1}{x}\right)-e^{\frac{1}{x}}\right]
$
Solution:
Let $y=\frac{1}{x}$ as $x \rightarrow 8, y \rightarrow 0$
$
\begin{aligned}
\lim _{x \rightarrow \infty} x\left[3^{\frac{1}{x}}+1-\cos \left(\frac{1}{x}\right)-e^{y / x}\right] & =\lim _{y \rightarrow 0} \frac{1}{y}\left(3^y+1-\cos y-e^y\right) \\
& =\lim _{y \rightarrow 0} \frac{3^y-1}{y}+\frac{1-\cos y}{y}-\frac{e^y+1}{y} \\
& =\lim _{y \rightarrow 0} \frac{3^y-1}{y}+\frac{2 \sin ^2 y / 2}{y}-\frac{\left(e^y-1\right)}{y} \\
& =\lim _{y \rightarrow 0}\left(\frac{3^y-1}{y}\right)+\frac{\sin ^2 \frac{y}{2}}{y / 2}-\frac{\left(e^y-1\right)}{y} \\
& =\lim _{y \rightarrow 0}\left(\frac{3^y-1}{y}\right)+\frac{\sin \frac{y}{2}}{y / 2} \times \sin (y / 2)-\left(\frac{e^y-1}{y}\right)
\end{aligned}
$

$
\begin{aligned}
& =\lim _{y \rightarrow 0}\left(\frac{3^y-1}{y}\right)+\lim _{y \rightarrow 0}\left(\frac{\sin ^y / 2}{y_2}\right) \lim _{y \rightarrow 0}(\sin y / 2)-\lim _{y \rightarrow 0} \frac{e^y-1}{y} \\
& =\log 3+(1)(0)-(1)=(\log 3)-1=-1+\log 3=\log 3-1
\end{aligned}
$
Question 19.
$
\lim _{x \rightarrow \infty}\{x[\log (x+a)-\log (x)]\}
$

Solution:
$
\begin{gathered}
\left.\lim _{x \rightarrow \infty}\{x \log (x+a)-\log x)\right\}=\lim _{x \rightarrow \infty} x\left[\log \left(\frac{x+a}{x}\right)\right] \\
=\lim _{x \rightarrow \infty} x[\log (1+a / x)]
\end{gathered}
$
Let $y=\frac{1}{x}$, Then as $x \rightarrow \infty, y \rightarrow \infty$
$
\begin{aligned}
& =\lim _{y \rightarrow \infty} \frac{1}{y} \log (1+a y) \\
& =\lim _{y \rightarrow 0} \frac{\log (1+a y)}{a y} \times a \\
& =a \lim _{y \rightarrow 0} \frac{\log (1+a y)}{a y} \\
& =a(1) \\
& =a
\end{aligned}
$

Question 20 .
$
\lim _{x \rightarrow \pi} \frac{\sin 3 x}{\sin 2 x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow \pi} \frac{\sin 3 x}{\sin 2 x} & =\lim _{x \rightarrow \pi} \frac{3 \sin x-4 \sin ^3 x}{2 \sin x \cos x}=\lim _{x \rightarrow \pi} \frac{\sin x\left(3-4 \sin ^2 x\right)}{\sin x(2 \cos x)} \\
& =\lim _{x \rightarrow \pi} \frac{3-4 \sin ^2 x}{2 \cos x}=\frac{3-4(0)}{2(-1)}=\frac{-3}{2}
\end{aligned}
$
Question 21.
$
\lim _{x \rightarrow \frac{\pi}{2}}(1+\sin x)^{2 \operatorname{cosec} x}
$
Solution:
let $y=\operatorname{cosec} x$, Then as $x \rightarrow \pi / 2, y \rightarrow \infty$
$
\lim _{x \rightarrow \pi / 2}(1+\sin x)^{2 \operatorname{cosec} x}=\lim _{y \rightarrow \infty}\left(1+\frac{1}{y}\right)^{2 y}=\lim _{y \rightarrow \infty}\left[\left(1+\frac{1}{y}\right)^y\right]^2=e^2
$

Question 22.
$
\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{2 \cos ^2 x} / 2}{\sin ^2 x} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{2} \cos x / 2}{\sin ^2 x} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{2}(1-\cos x / 2)}{\sin ^2 x}=\lim _{x \rightarrow 0} \frac{\sqrt{2} \times 2 \sin ^2 x / 4}{\sin ^2 x} \quad\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\
& =\lim _{x \rightarrow 0} 2 \sqrt{2}\left(\frac{\sin x / 4}{x / 4}\right)^2 \times\left(\frac{x}{\sin x}\right)^2 \times \frac{1}{16} \\
& =\frac{2 \sqrt{2}}{16} \lim _{x \rightarrow 0}\left(\frac{\sin x / 4}{x / 4}\right)^2 \times\left(\frac{1}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}\right)^2=\frac{2 \sqrt{2}}{16}(1)^2 \times \frac{1}{(1)^2}=\frac{\sqrt{2}}{8}=\frac{1}{4 \sqrt{2}} \\
&
\end{aligned}
$

Question 23.
$
\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \times\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right) \\
& =\lim _{x \rightarrow 0} \frac{(\not+\sin x)-(\not-\sin x)}{\tan x(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin x \times \cos x}{\sin x(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \\
& =\frac{2(1)}{1+1}=\frac{2}{2}=1 \\
&
\end{aligned}
$

Question 24.
$
\lim _{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^x
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^x \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{2 x-1}{x^2-4 x+2}\right)^x
\end{aligned}
$

$\begin{equation}
=\lim _{x \rightarrow \infty}\left(1+\frac{2 x-1}{x^2-4 x+2}\right)^{x\left(\frac{x^2-4 x+2}{2 x-1}\right) \times\left(\frac{2 x-1}{x^2-4 x+2}\right)}
\end{equation}$

$\begin{equation}
=\lim _{x \rightarrow \infty}\left(1+\frac{2 x-1}{x^2-4 x+2}\right)^{\left(\frac{x^2-4 x+2}{2 x-1}\right) \times \frac{x(2 x-1)}{\left(x^2-4 x+2\right)}}
\end{equation}$

Question 25 .
$
\lim _{x \rightarrow 0} \frac{e^x-e^{-x}}{\sin x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^x-e^{-x}}{\sin x}= & \lim _{x \rightarrow 0} \frac{e^x-\frac{1}{e^x}}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^x \sin x} \\
& =\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x e^x} \times\left(\frac{x}{\sin x}\right) \\
& =\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{2 x} \times 2 \times \lim _{x \rightarrow 0} \times \lim _{x \rightarrow 0} \frac{1}{\left(\frac{\sin x}{x}\right)} \\
& =(1)(2) \times 1 \times 1=2
\end{aligned}
$

Question 26.
$
\lim _{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^{a x}-e^{+b x}}{x} & =\lim _{x \rightarrow 0} \frac{e^{a x}-1-e^{b x}+1}{x} \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{a x}-1\right)-\left(e^{b x}-1\right)}{x} \\
& =\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x}-\left(\frac{e^{b x}-1}{x}\right) \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{a x}-1\right)}{a x} \times a-\lim _{x \rightarrow 0}\left(\frac{e^{b x}-1}{b x}\right) b=\log a-\log b \\
& =\log _b \frac{a}{b}
\end{aligned}
$
Question 27.
$
\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{x^3}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{x^3} & =\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{2 \sin ^2 x / 2}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{x^2}{2} \frac{\sin ^2 x / 2}{x^2 / 2} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{x^2}{2} \times \lim _{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)^2 \\
& =1 \times \frac{1}{2} \times 1=1 / 2
\end{aligned}
$

Question 28.
$
\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3} & =\lim _{x \rightarrow 0}\left(\frac{\frac{\sin x}{\cos x}-\sin x}{x^3}\right) \\
& \left.=\lim _{x \rightarrow 0} \frac{\sin x\left(\frac{1}{\cos x}-1\right)}{x^3}\right) \\
& =\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{\cos x \cdot x^3}=\lim _{x \rightarrow 0} \frac{\sin x 2 \sin ^2 x / 2}{\cos x \cdot x \cdot x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{x} \frac{2 \sin ^2 x / 2}{x^2} \times \frac{1}{\cos x} \\
& =\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right) \cdot\left(\frac{\sin x / 2}{x / 2}\right)^2 \times \frac{1}{2} \frac{1}{\cos x} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)^2 \cdot \lim _{x \rightarrow 0} \frac{1}{\cos x} \\
& =(1) \cdot 1 / 2 \cdot(1)=1 / 2
\end{aligned}
$