Exercise 9.5 - Chapter 9 Limits and Continuity 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$
\text { Ex } 9.5
$
Question 1.
Prove that $f(x)=2 x^2+3 x-5$ is continuous at all points in $R$. Solution:
Polynomial functions are continuous at every points of $\mathrm{R}$.
$
\lim _{x \rightarrow x_0} p(x)=\lim _{x \rightarrow x_0}\left(2 x^2+3 x-5\right)=2 x_0^2+3 x_0-5
$
Question 2.
Examine the continuity of the following:
(i) $x+\sin x$
Solution:
$f(x)=x+\sin x$
The Domain of the function $(-\infty, \infty)$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous in $(-\infty, \infty)($ i.e.,) for all $\mathrm{x} \in \mathrm{R}$
(ii) $x^2 \cos x$
Solution:
$f(x)=x^2 \cos x$
The Domain of the function $(-\infty, \infty)$
$
\begin{gathered}
\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} x^2 \cos x=c^2 \cos c \\
\lim _{x \rightarrow c^{+}} f(x)=\lim _{x \rightarrow c^{+}} x^2 \cos x=c^2 \cos c \\
\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)
\end{gathered}
$
$f(x)$ is continuous in $R$
(iii) $e^x \tan x$
Solution:
The Domain of the function in $\mathrm{R}-\{(2 \mathrm{n}+1) \pi / 2\}$
$\therefore$ The functions is continuous for all $\mathrm{x} \in \mathrm{R}-(2 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

(iv) $\mathrm{e}^{2 \mathrm{x}}+\mathrm{x}^2$
$
f(x)=e^{2 x}+x^2=1+2 x+\frac{(2 x)^2}{2 !}+\ldots \ldots \cdots \cdots+x^2
$
Solution:
$\therefore$ The functions is continuous for all $\mathrm{x} \in \mathrm{R}$
(v) $x \cdot \ln x$
Solution:
Thus $f(x)$ is continuous for $(0, \infty)$
(vi) $\frac{\sin x}{x^2}$
Solution:
Thus $f(x)$ is continuous for all $x \in \mathrm{R}-\{0\}$
(vii) $\frac{x^2-16}{x+4}$
Solution:
$
\mathrm{f}(\mathrm{x})=\frac{x^2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}
$
The function $f(x)$ is continuous for all $x \in R-\{-4\}$
(viii) $|\mathrm{x}+2|+|\mathrm{x}-1|$
Solution:
$f(x)$ is continuous for $x \in R$
(ix) $\frac{|x-2|}{|x+1|}$
Solution:
The function is continuous for all $x \in R-\{-1\}$
(x) $\cot x+\tan x$
Solution:
The function is continuous for all $\mathrm{x} \in \mathrm{R}-\frac{n \pi}{2}, \mathrm{n} \in \mathrm{z}$.

Question 3.
Find the points of discontinuity of the function $f$, where,
(i)
$
f(x)=\left\{\begin{array}{l}
4 x+5, \text { if } x \leq 3 \\
4 x-5, \text { if } x>3
\end{array}\right.
$
Solution:
$
f(3)=12+5=17
$
$
\begin{aligned}
& \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(4 x+5)=17 \\
& \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(4 x-5)=7 \\
& \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=3$
(ii)

$
f(x)= \begin{cases}x+2, & \text { if } x \geq 2 \\ x^2, & \text { if } x<2\end{cases}
$
Solution:
$
\begin{aligned}
& f(x)=4 \\
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(x+2)=4 \\
& \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} x^2=4
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous for all $\mathrm{x} \in \mathrm{R}$
(iii)
$
f(x)=\left\{\begin{array}{l}
x^3-3, \text { if } x \leq 2 \\
x^2+1, \text { if } x>2
\end{array}\right.
$
Solution:
$
\begin{aligned}
& f(x)=8-3=5 \\
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^3-3\right)=5 \\
& \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^2+1\right)=5
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous for all $\mathrm{x} \in \mathrm{R}$
(iv)
$
f(x)=\left\{\begin{array}{l}
\sin x, 0 \leq x \leq \frac{\pi}{4} \\
\cos x, \frac{\pi}{4}\frac{\pi}{2}
\end{array}\right.
$
Solution:

$
\begin{aligned}
f(\pi / 4) & =\sin \pi / 4=\frac{1}{\sqrt{2}} \\
\lim _{x \rightarrow \pi / 4^{+}} f(x) & =\lim _{x \rightarrow \pi / 4^{+}} \cos x=\frac{1}{\sqrt{2}}
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous for all $\mathrm{x} \in[0, \pi / 2]$
Question 4.
At the given points $\mathrm{x}_0$ discover whether the given function is continuous or discontinuous citing the reasons for your answer.
(i)
$
x_0=1, f(x)= \begin{cases}\frac{x^2-1}{x-1}, & x \neq 1 \\ 2, & x=1\end{cases}
$
Solution:
Given $\mathrm{f}\left(\mathrm{x}_0\right)=1$
$
\begin{aligned}
\lim _{x \rightarrow 1^{-}} f(x) & =\lim _{x \rightarrow 1^{-}} \frac{(x+1)(x-1)}{(x-1)}=2 \\
\lim _{x \rightarrow 1^{+}} f(x) & =2 \\
f(1) & =2
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}_0=1$
(ii)
$
x_0=3, f(x)= \begin{cases}\frac{x^2-9}{x-3}, & \text { if } x \neq 3 \\ 5, & \text { if } x=3\end{cases}
$
Solution:

$
\begin{aligned}
\lim _{x \rightarrow 3^{-}} f(x) & =\lim _{x \rightarrow 3^{-}} \frac{x^2-9}{(x-3)} \\
& =\lim _{x \rightarrow 3^{-}} \frac{(x+3)(x-3)}{(x-3)}=6
\end{aligned}
$
(i.e.,) $\quad \lim _{x \rightarrow 3^{-}} f(x)=6$
But $\quad \lim _{x \rightarrow 3^{+}} f(3)=5$
$
\therefore \quad \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(3)
$
$\therefore \mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}_0=3$
Question 5.
Show that :
$
\left\{\begin{array}{ll}
\frac{x^3-1}{x-1}, & \text { if } x \neq 1 \\
3, & \text { if } x=1
\end{array} \text { is continuous on }(-\infty, \infty)\right.
$

Solution
$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(\frac{x^3-1}{x-1}\right)=\lim _{x \rightarrow 1^{-}} \frac{(x-1)\left(x^2+x+1\right)}{(x-1)}=3
$
$
\lim _{x \rightarrow 1^{+}} f(x)=3
$
Given that $f(1)=3$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous for all $\mathrm{x} \in \mathrm{R}$
Question 6.
For what value of $\alpha$ is this function $\mathrm{f}(\mathrm{x})= \begin{cases}\frac{x^4-1}{x-1}, & \text { if } x \neq 1 \\ \alpha, & \text { if } x=1 \\ \text { continuous at } \mathrm{x}=1 \text { ? }\end{cases}$

Solution:

$
\begin{aligned}
\lim _{x \rightarrow 1} f(x) & =\lim _{x \rightarrow 1} \frac{x^4-1}{x-1} \\
& =\lim _{x \rightarrow 1} \frac{(x-1)(x+1)\left(x^2+1\right)}{(x-1)}=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)\left(x^2+1\right)}{(x-1)} \\
& =(2)(2) \\
& =4
\end{aligned}
$
$\because \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1, \alpha=4$

Question 7.
$
f(x)=\left\{\begin{array}{c}
0, \text { if } x<0 \\
x^2, \text { if } 0 \leq x<2 \\
4, \text { if } x \geq 2
\end{array}\right.
$

Solution:
$
f(x)=\left\{\begin{array}{cc}
0 & x<0 \\
x^2 & 0 \leq x<2 \\
4 & x \geq 2
\end{array}\right.
$
(i) $f(0)=0$
$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=0 \Rightarrow f(x) \text { is continuous at } x=0
$
(ii)
$
\begin{gathered}
f(2)=4 \\
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^2=4 \\
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(4)=4
\end{gathered}
$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous in $(-\infty, \infty)$
Question 8.
If $f$ and $g$ are continuous function with $f(3)=5$ and $\lim _{x \rightarrow 3}[2 f(x)-\boldsymbol{g}(x)]=4$, find $g(3)$. Solution:
Since $f$ and $g$ are continuous
$
\begin{aligned}
& \lim _{x \rightarrow 3}[2 f(x)-g(x)]=4 \\
& 2 f(3)-g(3)=4
\end{aligned}
$

$
\begin{aligned}
& 2(5)-g(3)=4 \\
& 10-4=g(3) \\
& g(3)=6
\end{aligned}
$
Question 9.
Find the points at which $f$ is discontinuous. At which of these points $f$ is continuous from the right, from the left, or neither? Sketch the graph of $\mathrm{f}$.
(i)
$
f(x)=\left\{\begin{array}{ccc}
2 x+1 & \text { if } & x \leq-1 \\
3 x & \text { if } & -1 2 x-1 & \text { if } & x \geq 1
\end{array}\right.
$
$\therefore \mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=1$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 1^{-}} f(x) & =\lim _{x \rightarrow 1^{-}}(3 x)=3 \\
\lim _{x \rightarrow 1^{+}} f(x) & =\lim _{x \rightarrow 1^{+}}(2 x-1)=2-1=1 \\
\lim _{x \rightarrow 1^{-}} f(x) & \neq \lim _{x \rightarrow 1^{+}} f(x)
\end{aligned}
$
$f(x)$ is not continuous at $x=1$
(ii)
$
f(x)=\left\{\begin{array}{l}
(x-1)^3, \text { if } x<0 \\
(x+1)^3, \text { if } x \geq 0
\end{array}\right.
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(x-1)^3=-1 \\
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x+1)^3=1 \\
& \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=0$

Question 10.
A function $f$ is defined as follows:
$
f(x)=\left\{\begin{array}{lll}
0 & \text { for } & x<0 \\
x & \text { for } & 0 \leq x<1 \\
-x^2+4 x-2 & \text { for } & 1 \leq x<3 \\
4-x & \text { for } & x \geq 3
\end{array}\right.
$
Is the function continuous?
Solution:

(i)
$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(x)=0
$
(ii)
$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x)=1
$
$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(-x^2+4 x-2\right)=-1+4-2=1
$
(iii)
$
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(-x^2+4 x-2\right)=-9+12-2=1
$
$
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(4-x)=4-3=1
$
From (i), (ii) and (iii)
$f(x)$ is continuous at $x=0,1,3$
Question 11.
Which of the following functions $\mathrm{f}$ has removable discontinuity at $\mathrm{x}=\mathrm{x}_0$ ? If the discontinuity is removable, find a function $g$ that agrees with $f$ for $x \neq x_0$ and is continuous on $R$.
(i) $\mathrm{f}(\mathrm{x})=\frac{x^2-2 x-8}{x+2}, \mathrm{x}_0=-2$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{x^2-2 x-8}{x+2}=\lim _{x \rightarrow 2} \frac{(x-4)(x+2)}{x+2}=-6 \\
& \therefore \quad g(x)=\left\{\begin{array}{cl}
\frac{x^2-2 x-8}{x+2} & \text { if } x \neq 2 \\
-6 & \text { if } x=-2
\end{array}\right.
\end{aligned}
$
(ii) $\mathrm{f}(\mathrm{x})=\frac{x^3+64}{x+4}, \mathrm{x}_0=-4$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 4} f(x)=\lim _{x \rightarrow 4} \frac{(x+4)\left(x^2-4 x+16\right)}{(x+4)}=48 \\
& \therefore \quad g(x)=\left\{\begin{array}{cl}
\frac{x^3+64}{x+4} & \text { if } x \neq-4 \\
48 & \text { if } x=-4
\end{array}\right.
\end{aligned}
$
(iii) $\mathrm{f}(\mathrm{x})=\frac{3-\sqrt{x}}{9-x}, \mathrm{x}_0=9$
Solution:
$f(x)=\frac{3-\sqrt{x}}{9-x}$, has a removable discontinuity at $x=9$
$
\begin{aligned}
\lim _{x \rightarrow 9} \frac{3-\sqrt{x}}{9-x} & =\lim _{x \rightarrow 9} \frac{3-\sqrt{x}}{3^2-(\sqrt{x})^2}=\lim _{x \rightarrow 9} \frac{(3-\sqrt{x})}{(3-\sqrt{x})(3+\sqrt{x})}=\frac{1}{6} \\
g(x) & =\left\{\begin{array}{cc}
\frac{3-\sqrt{x}}{9-x} & x \neq 9 \\
\frac{1}{6} & x=9 .
\end{array}\right.
\end{aligned}
$
Question 12.
Find the constant $\mathrm{b}$ that makes $g$ continuous on $(-\infty, \infty)$
$
g(x)=\left\{\begin{array}{l}
x^2-b^2, \text { if } x<4 \\
b x+20, \text { if } x \geq 4
\end{array}\right.
$
Solution:
Since $\mathrm{g}(\mathrm{x})$ is continuous,
$
\begin{aligned}
\lim _{x \rightarrow 4^{-}} g(x) & =\lim _{x \rightarrow 4^{+}} g(x) \\
\lim _{x \rightarrow 4^{-}}\left(x^2-b^2\right) & =\lim _{x \rightarrow 4^{+}} b x+20 \\
16-b^2 & =4 b+20 \\
b^2+4 b+4 & =0 \\
(b+2)^2 & =0
\end{aligned}
$

$
b=-2
$
Consider the function $f(x)=x \sin \frac{\pi}{x}$. What value must we give $f(0)$ in order to make the function continuous everywhere?
Solution:
$
f(x)=x \sin \frac{\pi}{x}
$
We know that $-1<\sin \frac{\pi}{x}<1$
$
\begin{gathered}
\text { so }-x \lim _{x \rightarrow 0} g(x)=\lim _{x \rightarrow 0}(-x)=0, \lim _{x \rightarrow 0} h(x)=\lim _{h \rightarrow 0}(x)=0 \\
\therefore \lim _{x \rightarrow 0} x \sin \frac{\pi}{x}=0
\end{gathered}
$
so to make the function $f(x)$ is continuous at $f(0)=0$

Question 14.
The function $f(x)=\frac{x^2-1}{x^3-1}$ is not defined at $x=1$. What value must we give $f(1)$ in order to make $f(x)$ continuous at $x=1$ ?
Solution:
$
\begin{gathered}
f(x)=\frac{x^2-1}{x^3-1}, f(x) \text { is not defined at } x=1 \\
\lim _{x \rightarrow 1} \frac{x^2-1}{x^3-1}=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{(x-1)\left(x^2+x+1\right)}=\frac{2}{3} \\
\therefore f(x)=\left\{\begin{array}{ccc}
\frac{x^2-1}{x^3-1} & \text { if } x \neq 1 \\
2 / 3 & \text { if } x=1 & \text { (i.e., } f(1)=\frac{2}{3}
\end{array}\right.
\end{gathered}
$
Question 15.
State how continuity is destroyed at $x=x_0$ for each of the following graphs.

Solution:
(a) $\lim _{x \rightarrow x_0^{-}} f(x) \neq \lim _{x \rightarrow x_0^{+}} f(x)$ is $f(x)$ is not continuous at $x=x_0$
(b) $f\left(x_0\right)$ is not defined as $f(x)$ is not continuous
(c) $f(x)$ is an vertical asymptote at $x=x_0$
(d) $\lim _{x \rightarrow x_0-} f(x) \neq \lim _{x \rightarrow x_0^{+}} f(x)$ is not continuous at $x=x_0$