Exercise 9.2 - Chapter 9 Limits and Continuity 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\operatorname{Ex} 9.2$
Evaluate the following limits:
Question 1.
$
\lim _{x \rightarrow 2} \frac{x^4-16}{x-2}
$
Solution:
$
\lim _{x \rightarrow 2} \frac{x^4-16}{x-2}=\lim _{x \rightarrow 2} \frac{x^4-2^4}{x-2}=4(2)^{4-1}=4 \times 8=32
$
Question 2.
$
\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}
$
Solution:
$
\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}=\lim _{x \rightarrow 1} \frac{\left(\frac{x^m-1}{x-1}\right)}{\left(\frac{x^n-1}{x-1}\right)}=\frac{\lim _{x \rightarrow 1}\left(\frac{x^m-1}{x-1}\right)}{\lim _{x \rightarrow 1} \frac{x^n-1^n}{x-1}}=\frac{m(1)^{m-1}}{n(1)^{n-1}}=\frac{m}{n}
$
Question 3.
$
\lim _{\sqrt{x} \rightarrow 3} \frac{x^2-81}{\sqrt{x}-3}
$
Solution:
$
\lim _{\sqrt{x} \rightarrow 3} \frac{x^2-81}{\sqrt{x}-3}=\lim _{\sqrt{x \rightarrow 3}} \frac{(\sqrt{x})^4-3^4}{(\sqrt{x})-(3)^1}=4(3)^3=4 \times 27=108
$

Question 4.
$
\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}, x>0
$
Solution:
$
\begin{aligned}
\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} & =\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\lim _{h \rightarrow 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\
& =\lim _{h \rightarrow 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2 \sqrt{x}}
\end{aligned}
$
Question 5.
$
\lim _{x \rightarrow 5} \frac{\sqrt{x+4}-3}{x-5}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 5} \frac{\sqrt{x+4}-3}{x-5} & =\lim _{x \rightarrow 5} \frac{\sqrt{x+4}-3}{x-5} \times \frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}=\lim _{x \rightarrow 5} \frac{(x+4)-9}{(x-5)(\sqrt{x+4}+3)} \\
& =\lim _{x \rightarrow 5} \frac{(x-5)}{(x-5)(\sqrt{x+4}+3)}=\frac{1}{\sqrt{9}+3}=\frac{1}{6}
\end{aligned}
$
Question 6.
$
\lim _{x \rightarrow 2} \frac{\frac{1}{x}-\frac{1}{2}}{x-2}
$
Solution:
$
\lim _{x \rightarrow 2} \frac{\frac{1}{x}-\frac{1}{2}}{x-2}=\lim _{x \rightarrow 2} \frac{x^{-1}-2^{-1}}{x-2}=(-1)(2)^{-2}=\frac{-1}{2^2}=\frac{-1}{4}
$
Question 7.
$
\lim _{x \rightarrow 1} \frac{\sqrt{x}-x^2}{1-\sqrt{x}}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{\sqrt{x}-x^2}{1-\sqrt{x}} & =\lim _{x \rightarrow 1} \frac{\sqrt{x}-x^2}{1-\sqrt{x}} \times \frac{1+\sqrt{x}}{1+\sqrt{x}} \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{x}+x-x^2-x^2 \sqrt{x}}{(1-x)} \\
& =\lim _{x \rightarrow 1} \frac{x(1-x)+\sqrt{x}\left(1-x^2\right)}{(1-x)} \\
& =\lim _{x \rightarrow 1} \frac{(1-x)[x+\sqrt{x}(1+x)]}{(1-x)} \\
& =1+(2)=3
\end{aligned}
$
Question 8 .
$
\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4} \times \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1} & \times \frac{\sqrt{x^2+16}+14}{\sqrt{x^2+16}+4} \\
& =\lim _{x \rightarrow 0} \frac{\left[\left(x^2+1\right)-1\right]\left[\sqrt{x^2+16}+4\right]}{\left(x^2+16-16\right)\left[\sqrt{x^2+1}+1\right]} \\
& =\lim _{x \rightarrow 0} \frac{x^2\left[\sqrt{x^2+16}+4\right]}{x^2\left[\sqrt{x^2+1}+1\right]} \\
& =\frac{4+4}{1+1}=\frac{8}{2}=4
\end{aligned}
$
Question 9.
$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\lim _{x \rightarrow 0} \frac{(1+x)^{1 / 2}-1^{1 / 2}}{(1+x)-1}= & \frac{1}{2}(1)^{\frac{1}{2}-1}=\frac{1}{2} \\
& (\text { OR) } \\
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}= & \lim _{x \rightarrow 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)} \\
= & \lim _{x \rightarrow 0} \frac{\not k}{x(\sqrt{1+x}+1)}=\frac{1}{2}
\end{aligned}
$
Question 10.
$
\lim _{x \rightarrow 1} \frac{\sqrt{7+x^3}-\sqrt{3+x^2}}{x-1}
$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(7+x^3\right)^{1 / 3}-\left(3+x^2\right)^{1 / 2}}{x-1} \\
= & \lim _{x \rightarrow 1} \frac{\left(7+x^3\right)^{1 / 3}-2-\left(3+x^2\right)^{1 / 2}+2}{x-1} \\
= & \lim _{x \rightarrow 1}\left(\frac{\left(7+x^3\right)^{1 / 3}-2}{x-1}\right)-\left(\frac{\left(3+x^2\right)^{1 / 2}-4^{1 / 2}}{x-1}\right) \\
= & \lim _{x \rightarrow 1} \frac{\left[\left(7+x^3\right)^{1 / 3}-8^{1 / 3}\right]\left[7+x^3-8\right]}{\left(7+x^3-8\right)(x-1)}-\frac{\left[\left(3+x^2\right)^{1 / 2}-4^{1 / 2}\right]\left[3+x^2-4\right]}{\left(3+x^2-4\right)(x-1)} \\
= & \lim _{x \rightarrow 1} \frac{\left[\left(7+x^3\right)^{1 / 3}-8^{1 / 3}\right]\left[x^3-1\right]}{\left(7+x^3-8\right)(x-1)}-\lim _{x \rightarrow 1} \frac{\left[\left(3+x^2\right)^{1 / 2}-4^{1 / 2}\right]\left[x^2-1\right]}{\left(3+x^2-4\right)(x-1)} \\
= & \lim _{x \rightarrow 1} \frac{\left[\left(7+x^3\right)^{1 / 3}-8^{1 / 3}\right]\left[(x-1)\left(n^2+x+1\right)\right]}{\left(7+x^3-8\right)(x-1)}-\lim _{x \rightarrow 1} \frac{\left[\left(3+x^2\right)^{1 / 2}-4^{1 / 2}\right](x-1)(n+1)}{\left(3+x^2-4\right)(x-1)} \\
= & \frac{1}{3}(8)^{\frac{1}{3}-1} \times 3-\frac{1}{2} \times(4)^{\frac{1}{2}-1} \times 2 \\
= & 8^{\frac{-2}{3}}-4^{\frac{-1}{2}} \\
= & \frac{1}{4}-\frac{1}{2} \\
= & \frac{-1}{4}
\end{aligned}
$
Question 11
$
\lim _{x \rightarrow 2} \frac{2-\sqrt{x+2}}{\sqrt[3]{2}-\sqrt[3]{4-x}}
$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{2-(x+2) \frac{1}{2}}{(2)^{1 / 3}-(4-x)^{1 / 3}} \\
& =\lim _{x \rightarrow 2} \frac{(x+2)^{1 / 2}-4^{1 / 2}}{(4-x)^{1 / 3}-2^{1 / 3}} \\
& =\lim _{x \rightarrow 2} \frac{\frac{(x+2)^{1 / 2}-(4)^{1 / 2}}{x+2-4}}{\frac{(4-x)^{1 / 3}-(2)^{1 / 3}}{(4-x)-2}} \times(-1) \\
& =-1 \times \frac{\lim _{x \rightarrow 2} \frac{(x+2)^{1 / 2}-(4)^{1 / 2}}{(x+2)-4}}{\lim _{x \rightarrow 2} \frac{(4-x)^{1 / 3}-(2)^{1 / 3}}{(4-x)-2}} \\
& =-1 \times \frac{\frac{1}{2}(4)^{1 / 2-1}}{\frac{1}{3}(2)^{1 / 3-1}}=-\frac{3}{2} \frac{(4)^{-1 / 2}}{(2)^{-2 / 3}} \\
& =-\frac{3}{2} \times \frac{\left[(2)^2\right]^{1 / 3}}{4^{1 / 2}}=\frac{-3}{2} \times \frac{\sqrt[3]{4}}{2} \\
& =\frac{-3}{2} \sqrt[3]{4} \\
&
\end{aligned}
$
Question 12.
$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-1}{x}
$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-1}{x} & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-1}{x} \times \frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1} \\
& =\lim _{x \rightarrow 0} \frac{\left(1+x^2\right)-1}{x\left(\sqrt{1+x^2}+1\right)}=\lim _{x \rightarrow 0} \frac{x^2}{x\left(\sqrt{1+x^2}+1\right)} \\
& =\frac{0}{2}=0
\end{aligned}
$
Question 13.
$
\lim _{x \rightarrow 0} \frac{\sqrt{1-x}-1}{x^2}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(\sqrt{1-x}-1)(\sqrt{1-x}+1)}{x^2(\sqrt{1-x}+1)} \\
& f(x) \rightarrow-\infty \text { as } x \rightarrow 0=\lim _{x \rightarrow 0} \frac{(1-x)-1}{x^2(\sqrt{1-x}+1)}=\lim _{x \rightarrow 0} \frac{-1}{x^2(\sqrt{1-x}-1)}
\end{aligned}
$
$\therefore$ Limit does not exist
Question 14 .
$
\lim _{x \rightarrow 5} \frac{\sqrt{x-1}-2}{x-5}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 5} \frac{\sqrt{x-1}-2}{x-5} \times \frac{\sqrt{x-1}+2}{\sqrt{x-1}+2} \\
& =\lim _{x \rightarrow 5} \frac{(x-1)-4}{x-5 \sqrt{x-1}+2}=\lim _{x \rightarrow 5} \frac{(x-5)}{(x-5)(\sqrt{x-1}+2)} \\
& =\frac{1}{2+2}=\frac{1}{4}
\end{aligned}
$

$
\lim _{\rightarrow a} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}(a>b)
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow:} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2} \\
& =\lim _{x \rightarrow a} \frac{(\sqrt{x-b}-\sqrt{a-b})(\sqrt{x-b}+\sqrt{a-b})}{\left(x^2-a^2\right)(\sqrt{x-b}+\sqrt{a-b})} \\
& =\lim _{x \rightarrow a} \frac{(x-b)-(a-b)}{(x-a)(x+a)(\sqrt{x-b}+\sqrt{a-b})} \\
& =\lim _{x \rightarrow a} \frac{x-b-a+b}{(x-a)(x+a)(\sqrt{x-b}+\sqrt{a-b})} \\
& =\lim _{x \rightarrow a} \frac{x-a}{(x-a)(x+a)(\sqrt{x-b}+\sqrt{a-b})} \\
& =\frac{1}{2 a(\sqrt{a-b}+\sqrt{a-b})} \\
& =\frac{1}{2 a(2 \sqrt{a-b})} \\
& =\frac{1}{4 a \sqrt{a-b}} \\
&
\end{aligned}
$