Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Put $x+1=y$ so that $y \to 1 $ as $x \to 0 $.

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

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

$=5 \cdot 1^{5-1}$

$[\lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}]$

$=5$

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

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{(x-2)(3 x+5)}{(x-2)(x+2)} \\ & =\lim _{x \to 2} \dfrac{3 x+5}{x+2} \\ & =\dfrac{3(2)+5}{2+2} \\ & =\dfrac{11}{4} \end{aligned} $

Answer :

At $x=3$, 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{(x-3)(x+3)(x^{2}+9)}{(x-3)(2 x+1)} \\ & =\lim _{x \to 3} \dfrac{(x+3)(x^{2}+9)}{2 x+1} \\ & =\dfrac{(3+3)(3^{2}+9)}{2(3)+1} \\ & =\dfrac{6 \times 18}{7} \\ & =\dfrac{108}{7} \end{aligned} $

Answer :

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

Answer :

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

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

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

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

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

$ =2.1^{2-1} \quad[\lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

$=2$

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

Answer :

$\lim _{x \to 1} \dfrac{a x^{2}+b x+c}{c x^{2}+b x+a}=\dfrac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}$

$ \begin{aligned} & =\dfrac{a+b+c}{a+b+c} \\ & =1 \quad[\because a+b+c \neq 0] \end{aligned} $

Answer :

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

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

Now, $\lim _{x \to-2} \dfrac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2}=\lim _{x \to-2} \dfrac{(\dfrac{2+x}{2 x})}{x+2}$

$ \begin{aligned} & =\lim _{x \to-2} \dfrac{1}{2 x} \\ & =\dfrac{1}{2(-2)}=\dfrac{-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, $\lim _{x \to 0} \dfrac{\sin a x}{b x}=\lim _{x \to 0} \dfrac{\sin a x}{a x} \times \dfrac{a x}{b x}$

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

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

$=\dfrac{a}{b} \times 1 \quad[\lim _{y \to 0} \dfrac{\sin y}{y}=1]$

$=\dfrac{a}{b}$

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

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

Answer :

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

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

$ \begin{aligned} \therefore \lim _{x \to \pi} \dfrac{\sin (\pi-x)}{\pi(\pi-x)} & =\dfrac{1}{\pi} \lim _{(\pi-x) \to 0} \dfrac{\sin (\pi-x)}{(\pi-x)} \\ & =\dfrac{1}{\pi} \times 1 \quad[\lim _{y \to 0} \dfrac{\sin y}{y}=1] \\ & =\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} \dfrac{x}{2}-1} \quad[\cos x=1-2 \sin ^{2} \dfrac{x}{2}] \\ & =\lim _{x \to 0} \dfrac{\sin ^{2} x}{\sin ^{2} \dfrac{x}{2}}=\lim _{x \to 0} \dfrac{(\dfrac{\sin ^{2} x}{x^{2}}) \times x^{2}}{(\dfrac{\sin ^{2} \dfrac{x}{2}}{(\dfrac{x}{2})^{2}}) \times \dfrac{x^{2}}{4}} \\ & =4 \dfrac{\lim _{x \to 0}(\dfrac{\sin ^{2} x}{x^{2}})}{\lim _{x \to 0}(\dfrac{\sin ^{2} \dfrac{x}{2}}{(\dfrac{x}{2})^{2}})} \\ & =4 \dfrac{(\lim _{x \to 0} \dfrac{\sin x}{x})^{2}}{(\lim _{\dfrac{x}{2} \to 0} \dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}})^{2}} \quad[x \to 0 \Rightarrow \dfrac{x}{2} \to 0] \\ & =4 \dfrac{1^{2}}{1^{2}} \quad[\lim _{y \to 0} \dfrac{\sin y}{y}=1] \\ & =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(a+\cos x)}{\sin x} \\ & =\dfrac{1}{b} \lim _{x \to 0}(\dfrac{x}{\sin x}) \times \lim _{x \to 0}(a+\cos x) \\ & =\dfrac{1}{b} \times \dfrac{1}{(\lim _{x \to 0} \dfrac{\sin x}{x})} \times \lim _{x \to 0}(a+\cos x) \\ & =\dfrac{1}{b} \times(a+\cos 0) \quad[\lim _{x \to 0} \dfrac{\sin x}{x}=1] \\ & =\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 :

$$ \begin{aligned} & \lim\limits_{x \to 0} \dfrac{sin ax+bx}{ax+sin bx} =\lim _{x \rightarrow 0} \frac{\sin (a x)+b x}{a x+\sin (b x)} \times \frac{b x}{a x} \times \frac{a}{b} \\ & =\lim _{x \rightarrow 0} \frac{\frac{\sin a x+b x}{a x}}{\frac{a x+\sin b x}{b x}} \times \frac{a}{b} \\ & =\frac{a}{b} \times \frac{\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x}}{\lim _{x \rightarrow 0} \frac{a x+\sin b x}{b x}} \\ & =\frac{a}{b} \times \frac{1+\frac{b}{a}}{1+\frac{a}{b}} \\ & =\frac{a}{b} \times \frac{b}{a} \\ & =1 \\ & \therefore \lim _{x \rightarrow 0} \frac{\sin (a x)+b x}{a x+\sin (b x)}=1 \end{aligned} $$

Answer :

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

Now,

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

Answer :

$\lim _{x \to \dfrac{\pi}{2}} \dfrac{\tan 2 x}{x-\dfrac{\pi}{2}}$

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

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

$ \begin{array}{ll} \therefore \lim _{x \to \dfrac{\pi}{2}} \dfrac{\tan 2 x}{x-\dfrac{\pi}{2}} & =\lim _{y \to 0} \dfrac{\tan 2(y+\dfrac{\pi}{2})}{y} & \\ & =\lim _{y \to 0} \dfrac{\tan (\pi+2 y)}{y} \quad [\tan (\pi+2 y)=\tan 2 y] \\ & =\lim _{y \to 0} \dfrac{\tan 2 y}{y} & \\ & =\lim _{y \rightarrow 0} \frac{\sin 2 y}{y \cos 2 y} \\ & =\lim _{y \to 0}(\dfrac{\sin 2 y}{2 y} \times \dfrac{2}{\cos 2 y}) & \\ & =(\lim _{2 y \to 0} \dfrac{\sin 2 y}{2 y}) \times \lim _{y \to 0}(\dfrac{2}{\cos 2 y}) \quad[\lim _{x \to 0} \dfrac{\sin x}{x}=1] \\ & =1 \times \dfrac{2}{\cos 0} & \\ & =1 \times \dfrac{2}{1} & \\ & =2 & \end{array} $

Answer :

The given function is

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

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

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

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

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

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

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

Answer :

The given function is

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

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

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

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

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

Answer :

The given function is

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

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

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

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

Answer :

The given function is

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

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

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

Answer :

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

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

$\therefore \lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=0$

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

Answer :

The given function is

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

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

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

$f(1)=4$

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

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

$\Rightarrow a+b=4$ 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 $f(x)=\left(x-a_1\right)(x-a_2) \ldots\left(x-a_n\right)$

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

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

Now, $\lim _{x \to a} f(x)=\lim _{x \to a}[(x-a_1)(x-a_2) \ldots(x-a_n)]$

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

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

Answer :

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

When $a=0$,

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

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

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

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

When $a<0$,

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

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

When $a>0$

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

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

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

Answer :

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

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

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

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

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

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

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

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

Answer :

The given function is

$ \begin{aligned} & f(x)= \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(x)=\lim _{x \to 0}(m x^{2}+n) \\ & =m(0)^{2}+n \\ & =n \\ & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0}(n x+m) \\ & =n(0)+m \\ & =m . \end{aligned} $

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

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

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

$ =n(1)^{3}+m $

$ =m+n $

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

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