Answer :

$\lim _{x \to 3} x+3=3+3=6$

Answer :

$\lim _{x \to \pi}\left(x-\dfrac{22}{7}\right)=\left(\pi-\dfrac{22}{7}\right)$

Answer :

$\lim _{r \to 1} \pi r^{2}=\pi\left(1\right)^{2}=\pi$

Answer :

$\lim _{x \to 4} \dfrac{4 x+3}{x-2}=\dfrac{4\left(4\right)+3}{4-2}=\dfrac{16+3}{2}=\dfrac{19}{2}$

Answer :

$\lim _{x \to-1} \dfrac{x^{10}+x^{5}+1}{x-1}=\dfrac{\left(-1\right)^{10}+\left(-1\right)^{5}+1}{-1-1}=\dfrac{1-1+1}{-2}=-\dfrac{1}{2}$

Answer :

$\lim _{x \to 0} \dfrac{\left(x+1\right)^{5}-1}{x}$

Put $x+1=y$ so that $y^5 - $ 1 as $x \to 0 $

Accordingly,

$\lim _{x \to 0} \dfrac{\left(x+1\right)^{5}-1}{x}=\lim _{y \to 1} \dfrac{y^{5}-1}{y-1}$

$\hspace{3cm}=\lim _{y \to 1} \dfrac{y^{5}-1^{5}}{y-1}$

$\hspace{3cm}=5 \cdot 1^{5-1}$ $\qquad\left[\because \ \ \lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]$

$\hspace{3cm}=5$

$\therefore \ \ \lim _{x \to 0} \dfrac{\left(x+5\right)^{5}-1}{x}=5$

Answer :

At $x=2$, the value of the given rational function takes the form $\dfrac{0}{0}$.

$ \begin{aligned} \therefore \ \ \lim _{x \to 2} \dfrac{3 x^{2}-x-10}{x^{2}-4} & =\lim _{x \to 2} \dfrac{\left(x-2\right)\left(3 x+5\right)}{\left(x-2\right)\left(x+2\right)} \\ \\ & =\lim _{x \to 2} \dfrac{3 x+5}{x+2} \\ \\ & =\dfrac{3\left(2\right)+5}{2+2} \\ \\ & =\dfrac{11}{4} \end{aligned} $

Answer :

At $x=2$, the value of the given rational function takes the form $\dfrac{0}{0}$.

$ \begin{aligned} \therefore \ \ \lim _{x \to 3} \dfrac{x^{4}-81}{2 x^{2}-5 x-3} & =\lim _{x \to 3} \dfrac{\left(x-3\right)\left(x+3\right)\left(x^{2}+9\right)}{\left(x-3\right)\left(2 x+1\right)} \\ \\ & =\lim _{x \to 3} \dfrac{\left(x+3\right)\left(x^{2}+9\right)}{2 x+1} \\ \\ & =\dfrac{\left(3+3\right)\left(3^{2}+9\right)}{2\left(3\right)+1} \\ \\ & =\dfrac{6 \times 18}{7} \\ \\ & =\dfrac{108}{7} \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{a x+b}{c x+1}=\dfrac{a\left(0\right)+b}{c\left(0\right)+1}=b$

Answer :

$\lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

At $z=1$, the value of the given function takes the form $\frac{0}{0}$

Put $z^{\frac{1}{6}}=x$ so that $z ^{\frac{1}{6}}- 1 \text{ as } x\to 1 $

Accordingly,

$\lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim _{x \to 1} \dfrac{x^{2}-1}{x-1}$

$ \hspace{2.2cm}=\lim _{x \to 1} \dfrac{x^{2}-1^{2}}{x-1} $

$ \hspace{2.2cm}=2.1^{2-1} \quad\left[\because \ \ \lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] $

$\hspace{2.2cm}=2$

$\therefore \ \ \lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$

Answer :

$ \begin{aligned} \lim _{x \to 1} \dfrac{a x^{2}+b x+c}{c x^{2}+b x+a} & =\dfrac{a\left(1\right)^{2}+b\left(1\right)+c}{c\left(1\right)^{2}+b\left(1\right)+a} \\ \\ & =\dfrac{a+b+c}{a+b+c} \\ \\ & =1 \quad\left[a+b+c \neq 0\right] \end{aligned} $

Answer :

$\lim _{x \to-2} \dfrac{\frac{1}{x}+\frac{1}{2}}{x+2}$

At $x=- $ 2 , the value of the given function takes the form $\frac{0}{0}$.

Now,

$ \begin{aligned} \lim _{x \to-2} \dfrac{\frac{1}{x}+\frac{1}{2}}{x+2} & =\lim _{x \to-2} \dfrac{\left(\frac{2+x}{2 x}\right)}{x+2} \\ \\ & =\lim _{x \to-2} \frac{1}{2 x} \\ \\ & =\frac{1}{2\left(-2\right)}=\frac{-1}{4} \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{\sin a x}{b x}$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$.

Now,

$\lim _{x \to 0} \dfrac{\sin a x}{b x}=\lim _{x \to 0} \dfrac{\sin a x}{a x} \times \dfrac{a x}{b x}$

$\hspace{2.2cm}=\lim _{x \to 0}\left(\dfrac{\sin a x}{a x}\right) \times\left(\dfrac{a}{b}\right)$

$\hspace{2.2cm}=\dfrac{a}{b} \lim _{a x \to 0}\left(\dfrac{\sin a x}{a x}\right) \quad\left[x \to 0 \Rightarrow a x \to 0\right]$

$\hspace{2.2cm}=\dfrac{a}{b} \times 1 \quad\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right]$

$\hspace{2.2cm}=\dfrac{a}{b}$

Answer :

$\lim _{x \to 0} \dfrac{\sin a x}{\sin b x}, a, b \neq 0$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$.

Now,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\sin a x}{\sin b x} & =\lim _{x \to 0} \dfrac{\left(\dfrac{\sin a x}{a x}\right) \times a x}{\left(\dfrac{\sin b x}{b x}\right) \times b x} \\ \\ & =\left(\dfrac{a}{b}\right) \times \dfrac{\lim _{a x \to 0}\left(\dfrac{\sin a x}{a x}\right)}{\lim _{b x \to 0}\left(\dfrac{\sin b x}{b x}\right)} \\ \\ & x \to 0 \Rightarrow a x \to 0 \ \text{ and } \ x \to 0 \Rightarrow b x \to 0 \\ \\ & =\left(\dfrac{a}{b}\right) \times \dfrac{1}{1} {\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right]} \\ \\ & =\dfrac{a}{b} \end{aligned} $

Answer :

$\lim _{x \to \pi} \dfrac{\sin \left(\pi-x\right)}{\pi\left(\pi-x\right)}$

It is seen that $x $ $\to$ $\pi \Rightarrow\left(\pi - x\right) \to 0 $

$ \begin{aligned} \therefore \ \ \lim _{x \to \pi} \dfrac{\sin \left(\pi-x\right)}{\pi\left(\pi-x\right)} & =\dfrac{1}{\pi} \lim _{\left(\pi-x\right) \to 0} \dfrac{\sin \left(\pi-x\right)}{\left(\pi-x\right)} \\ \\ & =\dfrac{1}{\pi} \times 1 \quad\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right] \\ \\ & =\dfrac{1}{\pi} \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{\cos x}{\pi-x}=\dfrac{\cos 0}{\pi-0}=\dfrac{1}{\pi}$

Answer :

$\lim _{x \to 0} \dfrac{\cos 2 x-1}{\cos x-1}$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$

Now,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\cos 2 x-1}{\cos x-1} & =\lim _{x \to 0} \dfrac{1-2 \sin ^{2} x-1}{1-2 \sin ^{2} \frac{x}{2}-1} \quad\left[\because \ \ \ \cos x=1-2 \sin ^{2} \dfrac{x}{2}\right] \\ \\ & =\lim _{x \to 0} \dfrac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}=\lim _{x \to 0} \dfrac{\left(\frac{\sin ^{2} x}{x^{2}}\right) \times x^{2}}{\left(\dfrac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right) \times \frac{x^{2}}{4}} \\ \\ & =4 \dfrac{\lim _{x \to 0}\left(\frac{\sin ^{2} x}{x^{2}}\right)}{\lim _{x \to 0}\left(\dfrac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right)} \\ \\ & =4 \dfrac{\left(\lim _{x \to 0} \frac{\sin x}{x}\right)^{2}}{\left(\lim _{\frac{x}{2} \to 0} \dfrac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}} \quad\left[x \to 0 \Rightarrow \dfrac{x}{2} \to 0\right] \\ \\ & =4 \left(\dfrac{1^{2}}{1^{2}}\right) \qquad\left[\because \ \ \ \lim _{y \to 0} \frac{\sin y}{y}=1\right] \\ \\ & =4 \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{a x+x \cos x}{b \sin x}$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$.

Now,

$ \begin{aligned} \lim _{x \to 0} \dfrac{a x+x \cos x}{b \sin x} & =\dfrac{1}{b} \lim _{x \to 0} \dfrac{x\left(a+\cos x\right)}{\sin x} \\ \\ & =\dfrac{1}{b} \lim _{x \to 0}\left(\dfrac{x}{\sin x}\right) \times \lim _{x \to 0}\left(a+\cos x\right) \\ \\ & =\dfrac{1}{b} \times \dfrac{1}{\left(\lim _{x \to 0} \dfrac{\sin x}{x}\right)} \times \lim _{x \to 0}\left(a+\cos x\right) \\ \\ & =\dfrac{1}{b} \times\left(a+\cos 0\right) \quad\left[\lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{a+1}{b} \end{aligned} $

Answer :

$\lim _{x \to 0} x \sec x=\lim _{x \to 0} \dfrac{x}{\cos x}=\dfrac{0}{\cos 0}=\dfrac{0}{1}=0$

Answer :

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$.

Now,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\sin a x+b x}{a x+\sin b x} & =\lim _{x \to 0} \dfrac{\left(\dfrac{\sin a x}{a x}\right) a x+b x}{a x+b x\left(\dfrac{\sin b x}{b x}\right)} \\ \\ & =\dfrac{\left(\lim _{a x \to 0} \dfrac{\sin a x}{a x}\right) \times \lim _{x \to 0}\left(a x\right)+\lim _{x \to 0} b x}{\lim _{x \to 0} a x+\lim _{x \to 0} b x\left(\lim _{b x \to 0} \dfrac{\sin b x}{b x}\right)} \qquad\left[\because \ \ \text{ As } x \to 0 \Rightarrow a x \to 0 \text{ and } b x \to 0 \right] \\ \\ & =\dfrac{\lim _{x \to 0}\left(a x\right)+\lim _{x \to 0} b x}{\lim _{x \to 0} a x+\lim _{x \to 0} b x} \qquad\left[\because \ \ \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{\lim _{x \to 0}\left(a x+b x\right)}{\lim _{x \to 0}\left(a x+b x\right)} \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{aligned} $

Answer :

At $x=0$, the value of the given function takes the form $\infty-\infty$.

Now,

$ \begin{aligned} \lim _{x \to 0}\left(\text{cosec} \ x-\cot x\right) & =\lim _{x \to 0}\left(\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{1-\cos x}{\sin x}\right) \\ \\ & =\lim _{x \to 0} \dfrac{\left(\dfrac{1-\cos x}{x}\right)}{\left(\dfrac{\sin x}{x}\right)} \\ \\ & =\dfrac{\lim _{x \to 0} \dfrac{1-\cos x}{x}}{\lim _{x \to 0} \dfrac{\sin x}{x}} \qquad\left[\because \ \ \lim _{x \to 0} \dfrac{1-\cos x}{x}=0 \text{ and } \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{0}{1} =0 \end{aligned} $

Answer :

$\lim _{x \to ({\pi}/{2})} \dfrac{\tan 2 x}{x-\frac{\pi}{2}}$

At $x=\dfrac{\pi}{2}$, the value of the given function takes the form $\dfrac{0}{0}$.

Now, put $x-\dfrac{\pi}{2}=y \quad x \to \dfrac{\pi}{2}, y \to 0$

$\therefore \ \ \lim _{x \to ({\pi}/{2})} \dfrac{\tan 2 x}{x-\frac{\pi}{2}} =\lim _{y \to 0} \dfrac{\tan 2\left(y+\frac{\pi}{2}\right)}{y} $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\tan \left(\pi+2 y\right)}{y} \quad\left[\because \ \ \tan \left(\pi+2 y\right)=\tan 2 y\right] $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\tan 2 y}{y} $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\sin 2 y}{y\cdot\cos 2 y} $

$\hspace{2.8cm} =\lim _{y \to 0}\left(\dfrac{\sin 2 y}{2 y} \times \dfrac{2}{\cos 2 y}\right) $

$\hspace{2.8cm} =\left(\lim _{2 y \to 0} \dfrac{\sin 2 y}{2 y}\right) \times \lim _{y \to 0}\left(\dfrac{2}{\cos 2 y}\right) \quad\left[\because \ \ \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] $

$\hspace{2.8cm} =1 \times \dfrac{2}{\cos 0} $

$\hspace{2.8cm} =1 \times \dfrac{2}{1} $

$\hspace{2.8cm} =2 $

Answer :

The given function is

$f\left(x\right)= \begin{cases}2 x+3, & x \leq 0 \\ \\ 3\left(x+1\right), & x>0\end{cases}$

$\lim _{x \to 0^{-}} f\left(x\right)=\lim _{x \to 0}\left(2 x+3\right)=2\left(0\right)+3=3$

$\lim _{x \to 0^{+}} f\left(x\right)=\lim _{x \to 0} 3\left(x+1\right)=3\left(0+1\right)=3$

$\therefore \ \ \lim _{x \to 0^{-}} f\left(x\right)=\lim _{x \to 0^{+}} f\left(x\right)=\lim _{x \to 0} f\left(x\right)=3$

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1} 3\left(x+1\right)=3\left(1+1\right)=6$ $\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} 3\left(x+1\right)=3\left(1+1\right)=6$

$\therefore \ \ \lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} f\left(x\right)=6$

Answer :

The given function is

$f\left(x\right)=\begin{cases} x^{2}-1, x \leq 1 \\ \\ -x^{2}-1, x>1 \end{cases} $

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1}\left(x^{2}-1\right)=1^{2}-1=1-1=0$

$\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1}\left(-x^{2}-1\right)=-1^{2}-1=-1-1=-2$

It is observed that $\lim _{x \to 1^{-}} f\left(x\right) \neq \lim _{x \to 1^{+}} f\left(x\right)$

Hence, $\lim _{x \to 1} f\left(x\right)$ does not exist.

Answer :

The given function is

$f\left(x\right)= \begin{cases}\dfrac{|x|}{x}, & x \neq 0 \\ \\ 0, & x=0\end{cases}$

$ \begin{matrix} \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(\dfrac{|x|}{x}\right) & \\ \\ & =\lim _{x \to 0}\left(\dfrac{-x}{x}\right) \quad & \\ \\ & =\lim _{x \to 0}\left(-1\right) \\ \\ & =-1 \\ \\ \lim _{x \to 0^{+}} f\left(x\right) & =\lim _{x \to 0^{+}}\left(\dfrac{|x|}{x}\right) & \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{x}\right) \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{matrix} $

It is observed that $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$.

Hence, $\lim _{x \to 0} f\left(x\right)$ does not exist.

Answer :

The given function is

$ \begin{aligned} f\left(x\right) & = \begin{cases}\dfrac{x}{|x|}, & x \neq 0 \\ \\ 0, & x=0\end{cases} \\ \\ \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(\dfrac{x}{|x|}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{-x}\right) \quad\left[\text{ When } x<0,|x|=-x\right] \\ \\ & =\lim _{x \to 0}\left(-1\right) \\ \\ & =-1 \\ \\ \lim _{x \to 0^{+}} f\left(x\right) & =\lim _{x \to 0^{+}}\left(\dfrac{x}{|x|}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{x}\right) \quad\left[\text{ When } x>0,|x|=x\right] \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{aligned} $

It is observed that $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$.

Hence, $\lim _{x \to 0} f\left(x\right)$ does not exist.

Answer :

The given function is $f\left(x\right)=|x|-5$.

$ \begin{aligned} & \begin{aligned} \lim _{x \to 5^{-}} f\left(x\right) & =\lim _{x \to 5^{-}}\left[|x|-5\right] \\ \\ & =\lim _{x \to 5}\left(x-5\right) \quad \\ \\ & =5-5 \\ \\ & =0 \end{aligned} \\ \\ & \begin{aligned} \lim _{x \to 5^{+}} f\left(x\right) & =\lim _{x \to 5^{+}}\left(|x|-5\right) \\ \\ & =\lim _{x \to 5}\left(x-5\right) \quad \\ \\ & =5-5 \\ \\ & =0 \end{aligned} \\ \\ & \therefore \ \ \lim _{x \to 5^{-}} f\left(x\right)=\lim _{x \to 5^{+}} f\left(x\right)=0 \end{aligned} $

Hence, $\lim _{x \to 5} f\left(x\right)=0$

Answer :

The given function is

$f\left(x\right)= \begin{cases}a+b x, & x<1 \\ \\ 4, & x=1 \\ \\ b-a x & x>1\end{cases}$

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1}\left(a+b x\right)=a+b$

$\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1}\left(b-a x\right)=b-a$

$f\left(1\right)=4$

It is given that $\lim _{x \to 1} f\left(x\right)=f\left(1\right)$.

$\therefore \ \ \lim _{x \to 1 ^{-}} f\left(x\right)=\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} f\left(x\right)=f\left(1\right)$

$\Rightarrow a+b=4$ and $b-a=4$

On solving these two equations, we obtain $a=0$ and $b=4$

Thus, the respective possible values of $a$ and $b$ are 0 and 4

Answer :

The given function is $f\left(x\right)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$

$ \begin{aligned} \lim _{x \to a_1} f\left(x\right) & =\lim _{x \to a_1}\left[\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right] \\ \\ & =\left[\lim _{x \to a_1}\left(x-a_1\right)\right]\left[\lim _{x \to a_1}\left(x-a_2\right)\right] \ldots\left[\lim _{x \to a_1}\left(x-a_n\right)\right] \\ \\ & =\left(a_1-a_1\right)\left(a_1-a_2\right) \ldots\left(a_1-a_n\right)=0 \end{aligned} $

$\therefore \ \ \lim _{x \to a_1} f\left(x\right)=0$

Now,

$ \begin{aligned} \lim _{x \to a} f\left(x\right) & =\lim _{x \to a}\left[\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right] \\ \\ & =\left[\lim _{x \to a}\left(x-a_1\right)\right]\left[\lim _{x \to a}\left(x-a_2\right)\right] \ldots\left[\lim _{x \to a}\left(x-a_n\right)\right] \\ \\ & =\left(a-a_1\right)\left(a-a_2\right) \ldots .\left(a-a_n\right) \end{aligned} $

$\therefore \ \ \lim _{x \to a} f\left(x\right)=\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right)$

Answer :

The given function is $f\left(x\right)= \begin{cases}|x|+1, & x<0 \\ \\ 0, & x=0 \\ \\ |x|-1, & x>0\end{cases}$

When $a=0$,

$ \begin{aligned} \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(|x|+1\right) \\ \\ & =\lim _{x \to 0}\left(-x+1\right) \quad\left[\text{ If } x<0,|x|=-x\right] \\ \\ & =(0+1) \\ \\ & =1 \end{aligned} $

$\lim _{x \to 0^{+}} f\left(x\right) =\lim _{x \to 0^{+}}\left(|x|-1\right) $

$\hspace{2.1cm} =\lim _{x \to 0}\left(x-1\right) \quad\left[\text{ If } x>0,|x|=x\right] $

$\hspace{2.1cm}=0-1 $

$\hspace{2.1cm} =-1 $

Here, it is observed that $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$

$\therefore \ \ \lim _{x \to 0} f\left(x\right)$ does not exist.

When $a<0$,

$ \begin{aligned} & \begin{aligned} \lim _{x \to a^{-}} f\left(x\right) & =\lim _{x \to a^{-}}\left(|x|+1\right) \\ \\ & =\lim _{x \to a}\left(-x+1\right) \quad\left[x<a<0 \Rightarrow|x|=-x\right] \\ \\ & =-a+1 \end{aligned} \\ \\ & \begin{aligned} \lim _{x \to a^{+}} f\left(x\right) & =\lim _{x \to a^{+}}\left(|x|+1\right) \\ \\ & =\lim _{x \to a}\left(-x+1\right) \quad\left[a<x<0 \Rightarrow|x|=-x\right] \\ \\ & =-a+1 \end{aligned} \\ \\ & \therefore \ \ \lim _{x \to a^{-}} f\left(x\right)=\lim _{x \to a^{+}} f\left(x\right)=-a+1 \end{aligned} $

Thus, limit of $f\left(x\right)$ exists at $x=a$, where $a<0$.

When $a>0$

$ \begin{aligned} \lim _{x \to a^{-}} f\left(x\right) & =\lim _{x \to a^{-}}\left(|x|-1\right) \\ \\ & =\lim _{x \to a}\left(x-1\right) \quad\left[0<x<a \Rightarrow|x|=x\right] \\ \\ & =a-1 \\ \\ \lim _{x \to a^{+}} f\left(x\right) & =\lim _{x \to a^{+}}\left(|x|-1\right) \\ \\ & =\lim _{x \to a}\left(x-1\right) \quad\left[0<a<x \Rightarrow|x|=x\right] \\ \\ & =a-1 \\ \\ & \therefore \ \ \lim _{x \to a^{-}} f\left(x\right)=\lim _{x \to a^{+}} f\left(x\right)=a-1 \end{aligned} $

Thus, limit of $f\left(x\right)$ exists at $x=a$, where $a>0$.

Thus, $\lim _{x \to a} f\left(x\right)$ exists for all $a \neq 0$

Answer :

$\lim _{x \to 1} \dfrac{f\left(x\right)-2}{x^{2}-1}=\pi$

$\Rightarrow \dfrac{\lim _{x \to 1}\left(f\left(x\right)-2\right)}{\lim _{x \to 1}\left(x^{2}-1\right)}=\pi$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=\pi \lim _{x \to 1}\left(x^{2}-1\right)$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=\pi\left(1^{2}-1\right)$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=0$

$\Rightarrow \lim _{x \to 1} f\left(x\right)-\lim _{x \to 1} 2=0$

$\Rightarrow \lim _{x \to 1} f\left(x\right)-2=0$

$\therefore \ \ \lim _{x \to 1} f\left(x\right)=2$

Answer :

The given function is

$ \begin{aligned} f\left(x \right) & = \begin{cases}m x^{2}+n, & x<0 \\ \\ n x+m, & 0 \leq x \leq 1 \\ \\ n x^{3}+m, & x>1\end{cases} \\ \\ \lim _{x \to 0^{-}} f\left(x \right) & =\lim _{x \to 0}\left(m x^{2}+n \right) \\ \\ & =m\left(0 \right)^{2}+n \\ \\ & =n \\ \\ \lim _{x \to 0^{+}} f\left(x \right) & =\lim _{x \to 0}\left(n x+m \right) \\ \\ & =n\left(0 \right)+m \\ \\ & =m . \end{aligned} $

Thus, $\lim _{x \to 0} f\left(x \right)$ exists if $m=n$.

$ \begin{aligned} \lim _{x \to 1^{-}} f\left(x \right) & =\lim _{x \to 1}\left(n x+m \right) \\ \\ & =n\left(1 \right)+m \\ \\ & =m+n \end{aligned} $

$\lim _{x \to 1^{+}} f\left(x \right)=\lim _{x \to 1}\left(n x^{3}+m \right)$

$ \hspace{2cm}=n\left(1 \right)^{3}+m $

$ \hspace{2cm}=m+n $

$\therefore \ \ \lim _{x \to 1^{-}} f\left(x \right)=\lim _{x \to 1^{+}} f(x)$

$\qquad\qquad\quad f\left(x \right)=\lim _{x \to 1} f\left(x \right)$

Thus, $\lim _{x \to 1} f\left(x \right)$ exists for any integral value of $m$ and $n$.