Exercise 9.3 - Chapter 9 Limits and Continuity 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 9.3
Question 1.
(a) Find the left right limits of $\mathrm{f}(\mathrm{x})=\frac{x^2-4}{\left(x^2+4 x+4\right)(x+3)}$ at $\mathrm{x}=-2$
(b) $f(x)=\tan x$ at $x=\frac{\pi}{2}$
Solution:
$
\begin{aligned}
& \text { (a) } \mathrm{f}(\mathrm{x})=\frac{x^2-4}{\left(x^2+4 x+4\right)(x+3)} \text { at } \mathrm{x} \rightarrow-2 \\
& =\lim _{x \rightarrow-2^{-}} \frac{x^2-4}{\left(x^2+4 x+4\right)(x+3)} \quad=\lim _{x \rightarrow-2^{-}} \frac{(x-2)(x+2)}{(x+2)^2(x+3)}=\lim _{x \rightarrow-2^{-}} \frac{x-2}{(x+2)(x+3)} \\
& \text { (i.e.) } f(-2) \rightarrow \infty \text { as } x \rightarrow-2- \\
& =\frac{-4}{-0}=\infty \\
& \lim _{x \rightarrow-2^{+}} \frac{x^2-4}{\left(x^2+4 x+4\right)(x+3)}=\lim _{x \rightarrow-2^{+}} \frac{(x-2)(x+2)}{(x+2)^2(x+3)}=\lim _{x \rightarrow-2^{-}} \frac{x-2}{(x+2)(x+3)} \\
& =\frac{-4}{+0}=-\infty \\
& \text { (b) } f(x)=\tan x \text { at } x=\pi / 2 \\
& \lim _{x \rightarrow \pi / 2} \tan x=\infty \\
& \text { (i.e.,) } f(\pi / 2) \rightarrow \infty \text { as } x \rightarrow \pi / 2^{-} \\
& \lim _{x \rightarrow \pi / 2^{+}} \tan x=-\infty \\
& \text { (i.e.,) } f(\pi / 2) \rightarrow-\infty \text { as } x \rightarrow \pi / 2^{+} \\
&
\end{aligned}
$
Evaluate the following limits
Question 2.
$
\lim _{x \rightarrow 5} \frac{x^2-9}{x^2\left(x^2-6 x+9\right)}
$

Solution:
Let $f(x)=\lim _{x \rightarrow 3} \frac{x^2-9}{x^2\left(x^2-6 x+9\right)}$
$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{x^2-9}{x^2\left(x^2-6 x+9\right)}=\lim _{x \rightarrow 3^{-}} \frac{(x-3)(x+3)}{x^2(x-3)^2}=\lim _{x \rightarrow 3^{-}} \frac{(x+3)}{x^2(x-3)}=-\infty \\
& f(3) \rightarrow-\infty \text { as } x \rightarrow 3^{-} \\
& \lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x^2\left(x^2-6 x+9\right)}=\lim _{x \rightarrow 3^{+}} \frac{(x-3)(x+3)}{x^2(x-3)^2}=+\infty \\
& f(3) \rightarrow \infty \text { as } x \rightarrow 3^{+}
\end{aligned}
$
Question 3.
$
\lim _{x \rightarrow \infty}-\frac{3}{x-2}-\frac{2 x+11}{x^2+x-6}
$
Solution:
$
\begin{aligned}
& \text { Lei } f(x)=\lim _{x \rightarrow \infty} \frac{3}{x-2}-\frac{2 x+11}{x^2+x-6} \\
& =\lim _{x \rightarrow \infty} \frac{3 x^2+3 x-18-(2 x+11)(x-2)}{(x-2)\left(x^2+x-6\right)} \\
& =\lim _{x \rightarrow \infty} \frac{\left(3 x^2+3 x-18\right)-\left(2 x^2-4 x+11 x-22\right)}{(x-2)\left(x^2+x-6\right)} \\
& \lim _{x \rightarrow \infty} \frac{x^2-4 x+4}{(x-2)\left(x^2+x-6\right)}=\lim _{x \rightarrow \infty} \frac{x^2\left(1-\frac{4}{x}+\frac{4}{x^2}\right)}{x^2\left(1+\frac{1}{x}-\frac{6}{x^2}\right)(x-2)} \\
& x^3+x \quad=\frac{1}{\infty}=0 \\
&
\end{aligned}
$
Question 4.
$
\lim _{x \rightarrow \infty} \frac{x^3+x}{x^4-3 x^2+1}
$

Question 7.
$
\lim _{x \rightarrow \infty}\left(\frac{x^3}{2 x^2-1}-\frac{x^2}{2 x+1}\right)
$
Solution:
Let $f(x)=\lim _{x \rightarrow \infty}\left(\frac{x^3}{2 x^2-1}-\frac{x^2}{2 x+1}\right)=\lim _{x \rightarrow \infty} \frac{x^3(2 x+1)-x^2\left(2 x^2-1\right)}{\left(2 x^2-1\right)(2 x+1)}$
$
\begin{aligned}
& =\lim _{x \rightarrow \infty} \frac{2 x^4+x^3-2 x^4+x^2}{4 x^3+2 x^2-2 x-1} \\
& =\lim _{x \rightarrow \infty} \frac{x^8\left(1+\frac{1}{x}\right)}{x^3\left(4+\frac{2}{x}-\frac{2}{x^2}-\frac{1}{x^3}\right)} \\
& =\frac{1}{4}
\end{aligned}
$
Question 8.
Show that (i) $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{3 n^2+7 n+2}=\frac{1}{6}$
(ii) $\lim _{n \rightarrow \infty} \frac{1^2+2^2+\ldots+(3 n)^2}{(1+2+\ldots+5 n)(2 n+3)}=\frac{9}{25}$
(iii) $\lim _{n \rightarrow \infty} \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots .+\frac{1}{n(n+1)}=1$

Solution:
(i)
$
\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{3 n^2+7 n+2}=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{3 n^2+7 n+2}=\lim _{n \rightarrow \infty} \frac{n^2\left(1+\frac{1}{n}\right)}{2 n^2\left(3+\frac{1}{n}+\frac{2}{n^2}\right)}=\frac{1}{6}
$
(ii)
$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+(3 n)^2}{(1+2+3+\ldots+5 n)(2 n+3)} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{(\not 3 n)(3 n+1)(6 n+1)}{6_2}}{\frac{(5 n)(5 n+1)}{2}(2 n+3)}=\lim _{n \rightarrow \infty} \frac{n^3\left(3+\frac{1}{n}\right)\left(6+\frac{1}{n}\right)}{5 n^3\left(5+\frac{5}{n}\right)\left(2+\frac{3}{n}\right)} \\
& =\frac{3 \times 63}{5 \times 5 \times 2}=\frac{9}{25} \\
&
\end{aligned}
$
(iii)
$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{n(n+1)} \\
& =\lim _{n \rightarrow \infty} \sum \frac{1}{n(n+1)}=\lim _{n \rightarrow \infty} \sum_{t=1}^n \frac{1}{n}-\frac{1}{n+1} \\
& =\lim _{n \rightarrow \infty}\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& =\lim _{n \rightarrow \infty} 1-\frac{1}{n+1}=1-0=1
\end{aligned}
$
Question 9.
An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or "recruits" that will return to the rivers during the reproductive period. If $\mathrm{S}$ is the number of spawners and $\mathrm{R}$ the number of recruits, "BevertonHolt spawner recruit function" is $\mathrm{R}(\mathrm{S})=\frac{S}{(\alpha S+\beta)}$ where $\alpha$ and $\beta$ are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.

Solution:
$
\begin{aligned}
& R(S)=\frac{S}{(\alpha S+\beta)} \\
& \text { as } S \rightarrow \infty, R(S)=\lim _{s \rightarrow \infty} \frac{S}{(\alpha S+\beta)} \\
& =\lim _{s \rightarrow \infty} \frac{S}{S\left(\alpha+\frac{\beta}{S}\right)}=\lim _{x \rightarrow \infty} \frac{1}{\left(\alpha+\frac{\beta}{S}\right)} \\
& =\frac{1}{\alpha} \\
&
\end{aligned}
$
Question 10 .
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after $\mathrm{t}$ minutes (in grams per litre) is $\mathrm{C}(\mathrm{t})=\frac{30 t}{200+t}$
What happens to the concentration as $\mathrm{t} \rightarrow \infty$ ?
Solution:
$
C(t)=\frac{30 t}{200+t}
$
as $t \rightarrow \infty, C(t)=\lim _{t \rightarrow \infty} \frac{30 t}{200+t}$
$
\begin{aligned}
& =\lim _{t \rightarrow \infty} \frac{30 t}{t\left(1+\frac{200}{t}\right)}=\frac{30}{1+0} \\
& =30
\end{aligned}
$