Exercise 9.1 - Chapter 9 - Limits and Continuity - 11th Maths Guide Samacheer Kalvi Solutions

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Limits and Continuity
Ex 9.1
In problems 1-6, complete the table using calculate and use the result to estimate the limit. Question 1.
$
\lim _{x \rightarrow 2} \frac{x-2}{x^2-x-2}
$
Solution:

Question 2.
$
\lim _{x \rightarrow 2} \frac{x-2}{x^2-4}
$
Solution:

Question 3.
$
\lim _{x \rightarrow 0} \frac{\sqrt{x+3}-\sqrt{3}}{x}
$
Solution:

Question 4.
$
\lim _{x \rightarrow 3} \frac{\sqrt{1-x}-2}{x+3}
$
Solution:

Question 5.
$
\lim _{x \rightarrow 0} \frac{\sin x}{x}
$
Solution:

Question 6.
$
\lim _{x \rightarrow 0} \frac{\cos x-1}{x}
$
Solution:

In exercise problems 7-15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
Question 7.
$
\lim _{x \rightarrow 3}(4-\mathrm{x})
$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 3}(4-x)=1 \\
& \lim _{x \rightarrow 3^{-}}(4-x)=1 \\
& \lim _{x \rightarrow 3^{+}}(4-x)=1 \\
& \lim _{x \rightarrow 3^{-}}(4-x)=\lim _{x \rightarrow 3^{+}}(4-x)=\lim _{x \rightarrow 3}(4-x)
\end{aligned}
$
Limit exist and is equal to 1
Question 8.
$
\lim _{x \rightarrow 1}\left(\mathrm{x}^2+2\right)
$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(x^2+2\right)=3 \\
& \lim _{x \rightarrow 1^{-}}\left(x^2+2\right)=3 \\
& \lim _{x \rightarrow 1^{+}}\left(x^2+2\right)=3 \\
& \lim _{x \rightarrow 1}\left(x^2+2\right)=\lim _{x \rightarrow 1^{-}}\left(x^2+2\right)=\lim _{x \rightarrow 1^{+}}\left(x^2+2\right)
\end{aligned}
$
Limit exist and is equal to $=3$
Question 9.
$\lim _{x \rightarrow 2} f(x)$ where $f(x)=\left\{\begin{array}{cc}4-x, & x \neq 2 \\ 0, & x=2\end{array}\right.$
Solution:

$
\begin{gathered}
\lim _{x \rightarrow 2} f(x) \\
\lim _{x \rightarrow 2^{-}}(4-x)=2 \\
\lim _{x \rightarrow 2^{+}}(4-x)=2 \\
\lim _{x \rightarrow 2^{-}}(4-x)=\lim _{x \rightarrow 2^{+}}(4-x)
\end{gathered}
$
Limit does exist $=2$
Question 10.
$\lim _{x \rightarrow 1} f(x)$ where $f(x)=\left\{\begin{array}{cc}x^2+2, & x \neq 1 \\ 1, & x=1\end{array}\right.$

Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 1} f(x)=1 \\
& \lim _{x \rightarrow 1^{-}}\left(x^2+2\right)=3 \\
& \lim _{x \rightarrow 1^{+}}\left(x^2+2\right)=3 \\
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\
& \Rightarrow \text { limt exist }=3
\end{aligned}
$
Question 11.
$
\lim _{x \rightarrow 3} \frac{1}{x-3}
$
Solution:
$
\begin{aligned}
& \lim _{x \rightarrow 3^{-}} \frac{1}{x-3}=-\infty \\
& \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty \\
& \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)
\end{aligned}
$
Limit does not exist

Question 12 .
$
\lim _{x \rightarrow 5} \frac{-|x-5|}{(x-5)}
$
Solution:

$
\begin{aligned}
& \text { When } \mathrm{x} \rightarrow 5,(\mathrm{x}-5)=-(\mathrm{x}-5) \\
& \therefore \lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)}=-1 \\
& \text { When } \mathrm{x} \rightarrow 5^{+},(\mathrm{x}-5)=(\mathrm{x}-5)
\end{aligned}
$

$
\begin{array}{rlrl} 
& \lim _{x \rightarrow 5^{+}} \frac{(x-5)}{(x-5)}=1 \\
\therefore \quad & \lim _{x \rightarrow 5^{-}} f(x) & \neq \lim _{x \rightarrow 5^{+}} f(x)
\end{array}
$
Limit does not exist
Question 13.
$\lim _{x \rightarrow 1} \sin (\pi x)$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 1} \sin (\pi x) \\
& \lim _{x \rightarrow 1^{-}} \sin (\pi x)=0 \\
& \lim _{x \rightarrow 1^{+}} \sin (\pi x)=0 \\
& \lim _{x \rightarrow 1} \sin (\pi x)=0 \\
& \lim _{x \rightarrow 1^{-}} \sin \pi x=\lim _{x \rightarrow 1^{+}} \sin \pi x=\lim _{x \rightarrow 1} \sin \pi x \\
& \text { So limit exists }=0
\end{aligned}
$
Question 14.
$
\lim _{x \rightarrow 0}(\sec x)
$

Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 0}(\sec x) \\
& \lim _{x \rightarrow 0^{-}}(\sec x)=1 \\
& \lim _{x \rightarrow 0^{+}} \sec x=1 \\
& \lim _{x \rightarrow 0} \sec x=1 \\
& \lim _{x \rightarrow 0^{-}} \sec x=\lim _{x \rightarrow 0^{+}} \sec x=\lim _{x \rightarrow 0} \sec x \\
& \text { limits exists } \\
& =1 \\
&
\end{aligned}
$

Question 15.
$\lim _{x \rightarrow \frac{\pi}{2}} \tan x$
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \tan x \\
& \lim _{x \rightarrow \frac{\pi^{-}}{2}} \tan x=-1 \\
& \lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan x=1 \\
& \lim _{x \rightarrow \frac{\pi}{2}^{-}} \tan x \neq \lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan x \\
&
\end{aligned}
$
Limit does not exist
Sketch the graph of $\mathrm{f}$, then identify the values of $\mathrm{x}_0$ for which $\lim _{x \rightarrow x_0} \mathrm{f}(\mathrm{x})$ exists.
Question 16.
$
f(x)= \begin{cases}x^2, & x \leq 2 \\ 8-2 x, & 2 $

Solution:

$
\begin{aligned}
\lim _{x \rightarrow 4^{-}}(8-2 x) & =0 \\
\lim _{x \rightarrow 4^{+}}(4) & =4 \\
\therefore \quad \lim _{x \rightarrow x_4} f(x) \text { does not exist at } x & =4 \\
\text { (i.e.) limit exists excepts at } x_0 & =4
\end{aligned}
$
Question 17.
$
f(x)= \begin{cases}\sin x, & x<0 \\ 1-\cos x, & 0 \leq x \leq \pi \\ \cos x, & x>\pi\end{cases}
$
Solution:

$
\begin{aligned}
\lim _{x \rightarrow 0^{-}} \sin x & =0, \quad \lim _{x \rightarrow 0^{+}} 1-\cos x=0 \\
\lim _{x \rightarrow \pi^{-}}(1-\cos x) & =\lim _{x \rightarrow \pi^{+}} \cos x=-1
\end{aligned}
$
Limit exists except at $x_0=\pi$
Question 18.
Sketch the graph of a function $f$ that satisfies the given values:
(i) $f(0)$ is undefined
(ii) $f(-2)=0$
$
\begin{array}{ll}
\lim _{x \rightarrow 0} f(x)=4 & f(2)=0 \\
f(2)=6 & \lim _{x \rightarrow 2} f(x) \\
\lim _{x \rightarrow 2} f(x)=3 & \lim _{x \rightarrow 2} f(x) \text { does not exist. }
\end{array}
$

Solution: 

(i)

(ii)

Question 19.
Write a brief description of the meaning of the notation $\lim _{x \rightarrow 8} f(x)=25$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow 8} f(x) & =25 \\
\lim _{x \rightarrow 8^{-}} f(x) & =25 \\
\lim _{x \rightarrow 8^{+}} f(x) & =25 \\
\lim _{x \rightarrow 8^{-}} f(x) & =\lim _{x \rightarrow 8^{+}} f(x) \\
f\left(8^{-}\right) & =f\left(8^{+}\right)=25 \text { (i.e.) } \lim _{x \rightarrow 8} f(x)=25
\end{aligned}
$
Question 20 .
If $f(2)=4$, can you conclude anything about the limit of $f(x)$ as $x$ approaches 2 ?
Solution:
No, $\mathrm{f}(\mathrm{x})=4$, It is the value of the function at $\mathrm{x}=2$
This limit doesn't exists at $\mathrm{x}=2$
Since $f(2)=4$
It need not imply that $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$
$\therefore$ we cannot conclude at $\mathrm{x}=2$
Question 21 .
If the limit of $f(x)$ as $z$ approaches 2 is 4 , can you conclude anything about $f(2)$ ?
Explain reasoning.
Solution:
$
\lim _{x \rightarrow 2^2} f(x) 4, \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=4
$
When $\mathrm{x}$ approaches 2 from the left or from the right $\mathrm{f}(\mathrm{x})$ approaches 4 .
Given that $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=4$
The existence or non existence at $x=2$ has no leaving on the existence of the limit of $f(x)$ as $x$ approaches to 2.
$\therefore$ We cannot conclude the value of $\mathrm{f}(2)$
Question 22 .
Evaluate: $\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}$ if it exists by finding $f\left(3^{-}\right)$and $f\left(3^{+}\right)$.
Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{x^2-9}{x-3} \\
& \lim _{x \rightarrow 3^{-}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{-}} \frac{(x+3)(x-3)}{(x-3)}=6 \\
& \lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{+}} \frac{(x+3)(x-3)}{(x-3)}=6 \\
& \lim _{x \rightarrow 3}\left(\frac{x^2-9}{x-3}\right)=6,6
\end{aligned}
$
Question 23.
Verify the existence of
$
\lim _{x \rightarrow 1} f(x), \text { where } f(x)=\left\{\begin{array}{r}
\frac{|x-1|}{x-1}, \text { for } x \neq 1 \\
0, \text { for } x=1
\end{array}\right.
$
Solution:
$
\lim _{x \rightarrow 1} f(x)
$
Where $\quad x \rightarrow 1^{-},|x-1|=-(x-1)$
$
\therefore \quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} \frac{-(x-1)}{x-1}=-1
$
$
\begin{aligned}
x \rightarrow 1^{+}-|x-1| & =(x-1) \\
\lim _{x \rightarrow 1^{+}} f(x) & =\lim _{x \rightarrow 1^{+}} \frac{(x-1)}{(x-1)}=1 \\
\lim _{x \rightarrow 1^{-}} f(x) & \neq \lim _{x \rightarrow 1^{+}} f(x)
\end{aligned}
$
Limit does not exist