# NCERT Solution of Class 11 Maths Chapter 12 Limit and Derivatives Exercise 12.1

In this page we have NCERT Solution of Class 11 Maths Chapter 12 Limit and Derivatives Exercise 12.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

## Evaluate the following limits

Question 1.
$\lim_{x \rightarrow 3} x + 3$
$\lim_{x \rightarrow 3} x + 3 = 3 + 3 = 6$

Question 2.
$\lim_{x \rightarrow \pi}\left ( x - \frac{22}{7} \right )$
$\lim_{x \rightarrow \pi} ( x - \frac{22}{7}) =( \pi - \frac{22}{7})$

Question 3.
$\lim_{r \rightarrow } \pi r^{2}$
$\lim_{r \rightarrow 1} \pi r^{2} = \pi (l)^{2} = \pi$

Question 4.
$\lim_{x \rightarrow 4}\frac{4x + 3}{x - 2}$
$\lim_{x \rightarrow 4}\frac{4x + 3}{x - 2} = \frac{4 (4) + 3}{4 - 2} = \frac{19}{2}$

Question 5.
$\lim_{x \rightarrow -1} \frac{x^{10} + x^{5} + 1}{x - 1}$
$\lim_{x \rightarrow -1} \frac{x^{10} + x^{5} + 1}{x - 1} = \frac{(-1)^{10} + (-1)^{5} + 1}{-1 - 1 } = \frac{1 - 1 + 1}{-2} = -\frac{1}{2}$

Question 6.
$\lim_{x \rightarrow 0}\frac{(x + 1)^{5} - 1}{x}$
$\lim_{x \rightarrow 0}\frac{(x + 1)^{5} - 1}{x}$
Put y=x+1 so that when x -> 0,y -> 1
$\lim_{y \rightarrow 1}\frac{y^{5} - 1}{y - 1} \\ = \lim_{y \rightarrow 1}\frac{y^{5} - 1^{5}}{y - 1}$
Now we know that
$\lim_{x \rightarrow a}\frac{x^{n} - a^{n}}{x - a}=na^{n-1}$$\lim_{y \rightarrow 1}\frac{y^{5} - 1^{5}}{y - 1} = 5.1^{5 - 1} \\ = 5$
Therefore,
$\lim_{x \rightarrow 0}\frac{(x + 1)^{5} - 1}{x} = 5$

Question 7
$\lim_{x \rightarrow 2}\frac{3x^{2} - x - 10}{x^{2} - 4}$
At x = 2, the given rational function has the value of 0/ 0, So we have to see if we can get the factor which is causing the terms to vanish
$\lim_{x \rightarrow 2}\frac{3x^{2} - x - 10}{x^{2} - 4}$
$= \lim_{x \rightarrow 2}\frac{ \left ( x - 2 \right ) \left ( 3x + 5 \right ) }{\left ( x - 2 \right ) \left ( x + 2 \right ) }$
$= \lim_{x \rightarrow 2} \frac{3x + 5}{x + 2}$ $= \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4}$

Question 8.
$\lim_{x \rightarrow 3}\frac{x^{4} - 81}{2x^{2} - 5x - 3}$
At x = 2, the given rational function has the value of 0/ 0,, So we have to see if we can get the factor which is causing the terms to vanish
$\lim_{x \rightarrow 3}\frac{x^{4} - 81}{2x^{2} - 5x - 3} = \lim_{x \rightarrow 3}\frac{(x- 3)(x + 3)(x^{2} + 9)}{(x - 3)(2x + 1)}$
$= \lim_{x \rightarrow 3}\frac{(x + 3)(x^{2} + 9)}{2x + 1}$
$= \frac{(3 + 3)(3^{2} + 9)}{2(3) + 1} = \frac{6 \times 18}{7}$
$= \frac{108}{7}$

Question 9.
$\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1}$ Answer
$\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0)+ 1} = b$

Question 10.
$\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$
$\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$
At z = 1, the given function has the value of 0/ 0
Put $z^{\frac{1}{6}} = x \; so \; that \; z\rightarrow 1 \; as \; x \rightarrow 1$
$Accordingly, \; \lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}} - 1}{z^{^{\frac{1}{6}}} - 1}$
$= \lim_{x \rightarrow 1} \frac{x^{2} - 1}{x - 1}$
$= \lim_{x \rightarrow 1}\frac{x^{2} - 1^{2}}{x - 1}$
$= 2.1^{2 - 1} \\ = 2$
Therefore,
$\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = 2$

Question 11.
$\lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a}, \; a + b + c \neq 0$
$\lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a} = \frac{a(1)^{2} + b(1) + c}{c(1)^{2} + b(1) + a}$ $= \frac{a + b + c}{a + b + c}$ = 1

Question 12.
$\lim_{x \rightarrow -2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$
At x = -2, the given function has the value of 0/ 0
$Now, \; \lim_{x \rightarrow -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \rightarrow -2} \frac{\left ( \frac{2 + x}{2x} \right )}{x + 2}$ $= \lim_{x \rightarrow -2}\frac{1}{2x}$ $= \frac{1}{2 (-2) } = \frac{-1}{4}$
Question 13.
$\lim_{x \rightarrow 0}\frac{\sin ax}{bx}$
At x = 0, the given function has the value of 0/ 0
$\\ Now, \lim_{x \rightarrow 0}\frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \frac{sin ax}{ax} \times \frac{ax}{bx}$ $= \lim_{x \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \times \left ( \frac{a}{b} \right )$ $= \frac{a}{b} \lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right )$ $= \frac{a}{b} \times 1 \\ = \frac{a}{b}$
Question 14.
$\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0$
At x = 0, the given function has the value of 0/ 0
$\\ Now, \lim_{x \rightarrow 0}\frac{sin ax}{sin bx} = \lim_{x \rightarrow 0} \frac { \left (\frac{ \sin ax}{ax} \right ) \times ax}{\left ( \frac{\sin bx}{bx} \right ) \times bx}\\ = \left ( \frac{a}{b} \right ) \times \frac{\lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right )}{\lim_{bx \rightarrow 0} \left ( \frac{\sin bx}{bx} \right )} \\ = \left ( \frac{a}{b} \right ) \times \frac{1}{1} \\ = \frac{a}{b}$
Question 15.
$\lim_{x \rightarrow \pi}\frac{\sin (\pi - x)}{\pi (\pi - x)}$
$\lim_{x \rightarrow \pi}\frac{\sin (\pi - x)}{\pi (\pi - x)}$
let $z=\pi - x$ now when x -> π,z -> 0
$\lim_{x \rightarrow \pi} \frac{\sin (\pi - x)}{\pi (\pi - x)}$
$= \frac{1}{\pi}\lim_{ z \rightarrow 0}\frac{\sin(z)}{(z)}$
$= \frac{1}{\pi} \times 1 \\ = \frac{1}{\pi}$

Question 16.
$\lim_{x \rightarrow 0}\frac{\cos x}{\pi - x}$
$\lim_{x \rightarrow 0}\frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi}$

Question 17.
$\lim_{x \rightarrow 0} \frac{\cos 2x - 1}{\cos x - 1}$
$\lim_{x \rightarrow 0} \frac{\cos 2x - 1}{\cos x - 1}$
At x = 0, the given function has the value of 0/ 0
Now, $\lim_{x \rightarrow 0}\frac{\cos 2x - 1}{\cos x - 1} = \lim_{x \rightarrow 0}\frac{1 - 2 \sin^{2} x - 1}{1 - 2 \sin^{2} \frac{x}{2} - 1}$
$= \lim_{x \rightarrow 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}} = \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin^{2}x}{x^{2}} \right ) \times x^{2}}{\left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )\times \frac{x^{2}}{4}}$
$= 4 \frac{\lim_{x\rightarrow 0}\left ( \frac{\sin^{2}x}{x^{2}} \right )}{\lim_{x \rightarrow 0} \left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )}$
$= 4 \frac{\left (\lim_{x \rightarrow 0}\frac{\sin x}{x} \right )^{2}}{\left ( \lim_{\frac{x}{2} \rightarrow 0} \frac{sin \frac{x}{2}}{\frac{x}{2}} \right )^{2}}$ $= 4\frac{1^{2}}{1^{2}}$
= 4

Question 18.
$\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}$
At x = 0, the given function has the value of 0/ 0
Now,
$\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x} = \frac{1}{b} \lim_{x \rightarrow 0}\frac{x(a + \cos x)}{\sin x}$
$= \frac{1}{b} \lim_{x \rightarrow 0} \left ( \frac{x}{\sin x} \right ) \times \lim_{x \rightarrow 0}(a + \cos x)$
$= \frac{1}{b} \times \frac{1}{\left ( \lim_{x \rightarrow 0} \frac{\sin x}{x} \right )} \times \lim_{x \rightarrow 0}\left ( a + \cos x \right )$
$= \frac{1}{b} \times \left ( a + \cos 0 \right ) \\ = \frac{a + 1}{b}$

Question 19.
$\lim_{x \rightarrow 0} x \sec x$
$\lim_{x \rightarrow 0} x \sec x = \lim_{x \rightarrow 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0$
Question 20.
$\lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} a, \;b \; \; a + b \neq 0$
At x =0, the given function has the value of 0/ 0
$= \frac{\left ( \lim_{ax \rightarrow 0} \frac{\sin ax}{ax} \right ) \times \lim_{x \rightarrow 0}\left ( ax \right ) + \lim_{x \rightarrow 0} bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0} bx\left ( \lim_{bx \rightarrow 0} \frac{\sin bx}{bx} \right )}$
$= \frac{\lim_{x \rightarrow 0}(ax) + \lim_{x \rightarrow 0}bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0}bx} \\ = \frac{\lim_{x \rightarrow 0}(ax + bx)}{\lim_{x \rightarrow 0}(ax + bx)}$
$= \lim_{x \rightarrow 0}(1)$
= 1

Question 21.
$\lim_{x \rightarrow 0}(\csc x - \cot x)$
At x =0, the given function has a undefined value
Now,
$\lim_{x \rightarrow 0}(\csc x - \cot x) \\ = \lim_{x \rightarrow 0}\left ( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right )$
$= \lim_{x \rightarrow 0}\left ( \frac{1 - \cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\frac{\left (\frac{1 - \cos x}{\sin x} \right )}{\left (\frac{\sin x}{x} \right )}$
$= \frac{\lim_{x \rightarrow 0}\frac{1 - \cos x}{x}}{\lim_{x \rightarrow 0}\frac{\sin x}{x}}$
Now we know that
$\lim_{x \rightarrow 0}\frac{\sin x}{x} =1$
and $\lim_{x \rightarrow 0}\frac{1 - \cos x}{x}=0$
$= \frac{0}{1}$
= 0

Question 22.
$\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x - \frac{\pi}{2}}$
At $x = \frac{\pi}{2}$, the given function has the value of 0/ 0
Now, put $x - \frac{\pi}{2} = y$ so that $x \rightarrow \frac{\pi}{2}, \; y \rightarrow 0$
Therefore,
$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\tan2x}{x - \frac{\pi}{2}} = \lim_{y \rightarrow 0}\frac{tan 2\left ( y + \frac{\pi}{2} \right ) }{y}$
$= \lim_{y \rightarrow 0} \frac{tan \left ( \pi + 2y \right ) }{y}$
$=\lim_{y \rightarrow 0} \frac{\tan 2y}{y}$
$= \lim_{y \rightarrow 0}\frac{\sin 2y}{y \cos 2y}$
$= \lim_{y \rightarrow 0}\left ( \frac{\sin 2y}{2y} \times \frac{2}{\cos 2y} \right )$
$= \left ( \lim_{2y \rightarrow 0} \frac{\sin 2y}{2y} \right ) \times \lim_{y \rightarrow 0}\left ( \frac{2}{\cos 2y} \right )$
Now we know that
$\lim_{x \rightarrow 0}\frac{\sin x}{x} =1$
$=1 \times \frac{2}{\cos 0} \\ = 1 \times \frac{2}{1}$
= 2

Question 23.
Find $\lim_{x\rightarrow 0} f(x)$ and $\lim_{x\rightarrow 1} f(x)$
where
$f(x) = \left\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\right.$
In these type of questions, we need to find the left hand and right hand limits and check if they are equal.Then we can say limit exist at that point
$f(x) = \left\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\right.$

$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0}\left [ 2x + 3 \right ] = 2(0) + 3 = 3$

$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0} 3(x + 1) = 3(0 + 1) = 3$

Therefore, $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0} f(x) = 3$

$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6$

$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6$

Therefore,
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} f(x)$ = 6

Question 24.
Find $\lim_{x \rightarrow 1} f(x)$ where
$f(x) = \left\{\begin{matrix} x^{2} - 1, & x \leq 1 \\ -x^{2} - 1, & x \geq 1 \end{matrix}\right.$
Given function is
$f(x) = \left\{\begin{matrix} x^{2} - 1, & x \leq 1 \\ -x^{2} - 1, & x \geq 1 \end{matrix}\right.$
In these type of questions, we need to find the left hand and right hand limits and check if they are equal.Then we can say limit exist at that point
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}[x^{2} - 1] = 1^{2} - 1 = 1 - 1 = 0$
$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1}[-x^{2} - 1] = -1^{2} - 1 = 1 - 1 = 0$
It is observed that $\lim_{x \rightarrow 1^{-}} f(x) \neq \lim_{x \rightarrow 1^{+}} f(x)$
Therefore, $\lim_{x \rightarrow 1} f(x)$ doesn't exist.

Question 25.
Find $\lim_{x \rightarrow 0} f(x)$ where
$f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.$
$f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.$
When x<0 then $\left | x \right | = -x$
$\lim_{x \rightarrow 0 ^{-}} f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{\left | x \right |}{x} \right ]$
$=\lim_{x \rightarrow 0 \left ( -1 \right )}$
= -1
When x >0 then $\left | x \right | = x$
$\lim_{x \rightarrow 0 ^{+}} f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{\left | x \right |}{x} \right ]$
$= \lim_{x \rightarrow 0 \left ( 1 \right )}$
= 1
It is observed that
$\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)$
Therefore, $\lim_{x \rightarrow 0 } f(x)$ doesn't exist.

Question 26.
Find $\lim_{x \rightarrow 0} f(x)$ where
$f(x) = \left\{\begin{matrix} \frac{x}{\left | x \right |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.$
$f(x) = \left\{\begin{matrix} \frac{x}{\left | x \right |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.$
When x<0 then $\left | x \right | = -x$
$\lim_{x \rightarrow 0^{-}}\left [ \frac{x}{ - x } \right ]$
$= \lim_{x \rightarrow 0^{-}}\left ( - 1 \right )$
= -1
When x >0 then $\left | x \right | = x$
$\lim_{x \rightarrow 0}\left [ \frac{x}{ x } \right ]$
$= \lim_{x \rightarrow 0}\left ( 1 \right )$
= 1
It is observed that
$\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)$
Therefore, $\lim_{x \rightarrow 0 } f(x)$ doesn't exist.

Question 27.
Find $\lim_{x \rightarrow 5} f(x)$ , where
$f(x) = \left | x \right | - 5$
Given function is $f(x) = \left | x \right | - 5$
When x > 0, $\left | x \right | = x$
$\lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}}[\left | x \right | - 5]$
$= \lim_{x \rightarrow 5} \left ( x - 5\right )$
= 5 - 5
= 0
When x > 0, $\left | x \right | = x$
$\lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}}(\left | x \right | - 5)$
$=\lim_{x \rightarrow 5^{+}} \left ( x - 5\right )$
= 5 - 5
= 0
Therefore,
$\lim_{x \rightarrow 5^{-}}f(x) = \lim_{x \rightarrow 5^{+}}f(x) = 0$
Hence, $\lim_{x \rightarrow 5 } f(x) = 0$

Question 28.
Suppose
$f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b - ax, & if \; x > 1 \end{matrix}\right.$
And
$\lim_{x \rightarrow 1} f(x) = f(1)$
what can be the values of b and a ?
Given function
$f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b - ax, & if \; x > 1 \end{matrix}\right.$
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(a + bx) = a + b$
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(b - ax) = b - a$
f(1) = 4
Given that $\lim_{x \rightarrow 1} f(x) = f(1)$.
Therefore,
$\lim_{x \rightarrow 1^{-}}f(x) = \lim_{x \rightarrow 1^{+}}f(x) = \lim_{x \rightarrow 1} f(x) = f(1)$ or
a+ b = 4 and b - a = 4
Solving both the equations, we get a = 0 and b = 4.
Hence, b and a are 4 and 0

Question 29.
A function $f(x) = (x - a_{1})(x - a_{2}) . . . . (x - a_{n})$ is define by fixed real numbers a1, a2, . . . . an.
Find $\lim_{x \rightarrow a_{1}} f(x)$ ?.
For some $a \neq a_{1}, a_{2}, . . . . , a_{n}$ , Compute $\lim_{x \rightarrow a_{n}} f(x)$
Given function
$\lim_{x \rightarrow a_{1}}f(x) = \lim_{x \rightarrow a_{1}}[(x - a_{1})(x - a_{2}) . . . (x - a_{n})]$

$= [\lim_{x \rightarrow a_{1}}(x - a_{1})][\lim_{x \rightarrow a_{1}}(x - a_{2})] . . . [\lim_{x \rightarrow a_{1}}(x - a_{n})]$

$= \left ( a_{1} - a_{1} \right ) \left ( a_{1} - a_{2} \right ) . . . . (a_{1} - a_{n}) = 0$
Therefore,
$\lim_{x \rightarrow a_{1}]}f(x) = 0$

Now,
$\lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a}[\left (x - a_{1} \right )\left ( x - a_{2} \right ) . . . \left ( x - a_{n} \right )]$

$= \lim_{x \rightarrow a}[ x - a_{1} ] [ x - a_{2}] . . . [x - a_{n}]$

=$( a - a_{1}) ( a - a_{2} ) . . . . (a - a_{n})$
Therefore,
$\lim_{x \rightarrow a} f(x) =( a - a_{1})( a - a_{2}) . . . . (a - a_{n})$

Question 30.
If $f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | - 1, & x > 0 \end{matrix}\right.$
For what value (s) of a does $\lim_{x \rightarrow a} f(x)$ limit exists?
Given function:
$f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | - 1, & x > 0 \end{matrix}\right.$
First lets check the limit at a = 0
Now when x < 0, |x| = -x
$\lim_{x \rightarrow 0^{-}}\left ( - x + 1 \right )$
= -0 + 1
= 1
Now when x > 0, |x| = x
$\lim_{x \rightarrow 0}\left ( x - 1 \right )$
= 0 - 1
= -1
We have, $\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)$
Therefore, $\lim_{x \rightarrow 0} f(x)$ doesn't exist.
Now lets check the limit for values of a < 0
$\lim_{x \rightarrow a^{-}}f(x) = \lim_{x \rightarrow a^{-}}(\left | x \right | + 1)$
$= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = - x ]$
= -a + 1
$\lim_{x \rightarrow a^{+}}f(x) = \lim_{x \rightarrow a^{+}}(\left | x \right | + 1)$
$= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ a < x < 0 \Rightarrow \left | x \right | = - x ]$
= - a + 1
$\lim_{x \rightarrow a^{-} } = \lim_{x \rightarrow a^{+} } = - a + 1$
Hence, at x = a limit of f(x) exist, where a < 0
Now lets check the limit for values of a > 0
$\lim_{x \rightarrow a^{-}}f(x) = \lim_{x \rightarrow a^{-}}(\left | x \right | + 1)$
$= \lim_{x \rightarrow a}( x - 1) \; \; \; \; [ 0 < x < a \Rightarrow \left | x \right | = x ]$
= a - 1
$\lim_{x \rightarrow a^{+}}f(x) = \lim_{x \rightarrow a^{+}}(\left | x \right | + 1)$
$= \lim_{x \rightarrow a}( x - 1) \; \; \; \; [ 0 < a < x \Rightarrow \left | x \right | = x ]$
= a - 1
Hence, at x = a ,limit of f(x) exist, where a > 0
Thus, $\lim_{x \rightarrow a} f(x) \; exist \; for \; all \; a \neq 0$

Question 31.
If f(x) satisfies, $\lim_{x \rightarrow 1} \frac{f(x) - 2}{x^{2} - 1} = \pi$
then Find $\lim_{x \rightarrow 1} f(x)$
$\Rightarrow \lim_{x \rightarrow 1} (f(x) - 2) = \pi \lim_{x \rightarrow 1} (x^{2} - 1)$
$\Rightarrow \lim_{x \rightarrow 1} (f(x) - 2) = 0$
$\Rightarrow \lim_{x \rightarrow 1} f(x) - 2 = 0$
$\lim_{x \rightarrow 1} f(x) = 2$

Question 32.
If $f(x) = \left\{\begin{matrix} mx^{2} + n, & x < 0 \\ mx + n, & 0 \leq x \leq 1 \\ nx^{3} + m & x > 1 \end{matrix}\right.$
For what values of m and n does $\lim_{x \rightarrow 0} f(x)$ and $\lim_{x \rightarrow 1} f(x)$ exist?

Given function
$f(x) = \left\{\begin{matrix} mx^{2} + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^{3} + m & x > 1 \end{matrix}\right.$
$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0}(mx^{2} + n)$
= m(0)2 + n
= n
$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0}(nx + m)$
= n(0) + m
= m
Therefore, $\lim_{x \rightarrow 0} f(x)$ will exists if m=n
Now
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(nx + m)$
= n(1) + m
= m+n
$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1}(nx^{3} + m)$
= n + m
Therefore,
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} f(x)$ Thus, for integral values of n and m $\lim_{x \rightarrow 1}f(x)$ exist.

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