Limits, Continuity and Differentiability · Mathematics · MHT CET

$$\lim _\limits{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}=$$

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by

$$f(x)=\left\{\begin{array}{cc} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k \quad, & x=\frac{\pi}{4} \end{array}\right.$$

is continuous, then k is equal to

The value of $\lim _\limits{x \rightarrow 0}\left((\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$, where $x>0$ is

Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is

If the function $f(x)= \begin{cases}-2 \sin x & \text {, if } x \leq \frac{-\pi}{2} \\ A \sin x+B & , \text { if } \frac{-\pi}{2}< x<\frac{\pi}{2} \\ \cos x & , \text { if } x \geq \frac{\pi}{2}\end{cases}$ is continuous everywhere, then the values of $A$ and B are respectively

$$\lim _\limits{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}=$$

Let $f:[-1,3] \rightarrow \mathbb{R}$ be defined as

$$\left\{\begin{array}{lc} |x|+[x], & -1 \leqslant x<1 \\ x+|x|, & 1 \leqslant x<2 \\ x+[x], & 2 \leqslant x \leqslant 3 \end{array}\right.$$

where $[t]$ denotes the greatest integer function. Then $f$ is discontinuous at

$$\lim _\limits{x \rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$$ is

$\lim _\limits{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$ is

Let $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , x<0 \\ a & , x=0 \\ \frac{\sqrt{2}}{\sqrt{16+\sqrt{x-4}}} & , x>0\end{array}\right.$ If $\mathrm{f}(x)$ is continuous at $x=0$, then the value of $a$ is

Let f be twice differentiable function such that $\mathrm{f}^{\prime \prime}(x)=-\mathrm{f}(x), \mathrm{f}^{\prime}(x)=\mathrm{g}(x)$ and $\mathrm{h}(x)=(\mathrm{f}(x))^2+(\mathrm{g}(x))^2$. If $\mathrm{h}(5)=1$, then the value of $h(10)$ is

$$\lim _\limits{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$$

If $f(x)=\left\{\begin{array}{cc}\frac{a}{2}(x-|x|) & , \\ 0, & \text { for } x<0 \\ 0, & \text { for } x=0 \\ b x^2 \sin \left(\frac{1}{x}\right) & \text { for } x>0\end{array}\right.$

is continuous at $x=0$, then

The approximate value of $(3.978)^{\frac{3}{2}}$ is

If $\mathrm{f}(x)=\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is continuous at $x=0$ then $f(0)$ is equal to

$$\lim _\limits{x \rightarrow 0} \frac{9^x-4^x}{x\left(9^x+4^x\right)}=$$

The values of $a$ and $b$, so that the function

$$f(x)= \begin{cases}x+\mathrm{a} \sqrt{2} \sin x & , 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b & , \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ \mathrm{a} \cos 2 x-\mathrm{b} \sin x & , \frac{\pi}{2}< x \leq \pi\end{cases}$$

is continuous for $0 \leq x \leq \pi$, are respectively given by

The approximate value of $x^3-2 x^2+3 x+2$ at $x=2.01$ is

The approximate value of $(3.978)^{3 / 2}$ is

$$\lim _\limits{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(8 x^3-\pi^3\right) \cos x}{(\pi-2 x)^4}$$

Let $\mathrm{f}(x)=x\left[\frac{x}{2}\right]$, for $-10< x<10$, where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

If $\lim\limits_{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$ then

Let k be a non-zero real number. If $f(x)=\left\{\begin{array}{cl}\frac{\left(\mathrm{e}^x-1\right)^2}{\sin \left(\frac{x}{k}\right) \log \left(1+\frac{x}{4}\right)} & , x \neq 0 \\ 12 & , x=0\end{array}\right.$ is a continuous function, then the value of $k$ is

If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)=$

If $\lim _\limits{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=7$, then $a+b$ is equal to

If $f(x)=\left(\sin ^4 x+\cos ^4 x\right), 0< x<\frac{\pi}{2}$, then the function has minimum value at $x=$

$\lim _\limits{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ has the value

Let $a, b \in(a \neq 0)$. If the function $f$ is defined as

$$f(x)=\left\{\begin{array}{cc} \frac{2 x^2}{\mathrm{a}} & , 0 \leq x<1 \\ \mathrm{a} & , 1 \leq x<\sqrt{2} \\ \frac{2 \mathrm{~b}^2-4 b}{x} & , \sqrt{2} \leq x<\infty \end{array}\right.$$

is continuous in the interval $[0, \infty)$, then an ordered pair $(a, b)$ is

If the function $\mathrm{f}(x)=\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}$ if $x \neq 2$. $=\mathrm{k}$ if $x=2$. is continuous at $x=2$, then $\mathrm{k}=$

For each $x \in \mathbb{R}$, Let $[x]$ represent greatest integer function, then $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to

Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{1}{2}\right], \quad \mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $\mathrm{f}\left(\frac{\pi}{4}\right)$ is

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $$(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$$ where $a>-1$ then $\lim _\limits{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _\limits{a \rightarrow 0^{+}} \beta(a)$ respectively are

The value of k , for which the function

$$\mathrm{f}(x)= \begin{cases}\left(\frac{4}{5}\right)^{\frac{\ln 4 x}{\tan 5 x}}, & 0< x< \frac{\pi}{2} \\ \mathrm{k}+\frac{2}{5} & , x=\frac{\pi}{2}\end{cases}$$

is continuous at $x=\frac{\pi}{2}$, is

$\lim _\limits{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to

The value of $\lim _\limits{x \rightarrow 0} \frac{x}{|x|+x^2}$ is

If $\mathrm{f}(x)=\frac{1+\cos \pi x}{\pi(1-x)^2}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)$ is equal to

Let K be the set of all real values of $x$, where the function $\mathrm{f}(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then the set K is

The function $\mathrm{f}$ defined on $$\left(-\frac{1}{3}, \frac{1}{3}\right)$$ by $$\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , \quad x \neq 0 \\ \mathrm{k} & , \quad x=0 \end{array}\right.$$ is continuous at $$x=0$$, then $$\mathrm{k}$$ is

If $$f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$$, then as $$x$$ approaches a, $$\frac{\mathrm{g}(x) \mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a}) \mathrm{f}(x)}{(x-\mathrm{a})}$$ approaches

Let $$\mathrm{f}(x)=5-|x-2|$$ and $$\mathrm{g}(x)=|x+1|, x \in \mathrm{R}$$ If $$\mathrm{f}(x)$$ attains maximum value at $$\alpha$$ and $$\mathrm{g}(x)$$ attains minimum value at $$\beta$$, then $$\lim _\limits{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$$ is equal to

$$\lim _\limits{x \rightarrow \infty} x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\}=$$

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}=$$

If $$f(x)$$ is continuous on its domain $$[-2,2]$$, where

$$f(x)=\left\{\begin{array}{cc} \frac{\sin a x}{x}+3 & , \text { for }-2 \leq x<0 \\ 2 x+7 & , \text { for } 0 \leq x \leq 1 \\ \sqrt{x^2+8}-b & , \text { for } 1< x \leq 2 \end{array}\right.$$ $$\text { then the value of } 2 a+3 b \text { is }$$

The function $$\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$$ where $$\mathrm{t}=\frac{1}{x-1}$$ is discontinuous at

If $$\mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7$$, then $$\lim _\limits{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}=$$

$$\lim _\limits{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2(2 x)} \text { is equal to }$$

Given $$\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , \text { if } x<0 \\ \mathrm{a} & , \text { if } x=0 \\ \frac{\sqrt{x}}{\sqrt{16-\sqrt{x}-4}}, & \text { if } x>0\end{array}\right.$$

If $$\mathrm{f}(x)$$ is continuous at $$x=0$$, then value of a is

The values of $$a$$ and $$b$$, so that the function

$$f(x)=\left\{\begin{array}{l} x+a \sqrt{2} \sin x, 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, \frac{\pi}{2} < x \leq \pi \end{array}\right.$$

is continuous for $$0 \leq x \leq \pi$$, are respectively given by

$$\lim _\limits{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2}$$ is

$$\text { If } l=\lim _\limits{x \rightarrow 0} \frac{x}{|x|+x^2} \text {, then the value of } l \text { is }$$

If $$\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{x-3}{|x-3|}+\mathrm{a} & , \quad x < 3 \\ \mathrm{a}+\mathrm{b} & , \quad x=3 \\ \frac{|x-3|}{x-3}+\mathrm{b}, & x>3\end{array}\right.$$

Is continuous at $$x=3$$, then the value of $$\mathrm{a}-\mathrm{b}$$ is

$$\lim _\limits{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$$ is equal to

The left-hand derivative of $$\mathrm{f}(x)=[x] \sin (\pi x)$$, at $$x=\mathrm{k}, \mathrm{k}$$ is an integer and [.] is the greatest integer function, is

If $$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}, & -1 \leq x < 0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{array}\right.$$ is continuous in the interval $$[-1,1]$$, then $$\mathrm{m}$$ is equal to

Let $$\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R} / \mathrm{f}(x)=|x-\pi|\left(\mathrm{e}^{|x|}-1\right) \sin |x|\right.$$ is not differentiable at $$\mathrm{t}\}$$, then $$\mathrm{S}$$ is

If $$\mathrm{f}(x)=\frac{4}{x^4}\left[1-\cos \frac{x}{2}-\cos \frac{x}{4}+\cos \frac{x}{2} \cdot \cos \frac{x}{4}\right]$$ is continuous at $$x=0$$, then $$\mathrm{f}(0)$$ is

$$\lim _\limits{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}=$$

The value of $$\lim _\limits{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$$ is

If the function $$\mathrm{f}(x)$$ is continuous in $$0 \leq x \leq \pi$$, then the value of $$2 a+3 b$$ is where

$$f(x)= \begin{cases}x+a \sqrt{2} \sin x & \text { if } 0 \leq x < \frac{\pi}{4} \\ 2 x \cot x+b & \text { if } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ \operatorname{acos} 2 x-b \sin x & \text { if } \frac{\pi}{2} < x \leq \pi\end{cases}$$

$$\lim _\limits{x \rightarrow 0} \frac{(1-\cos 2 x) \cdot \sin 5 x}{x^2 \sin 3 x}$$ is

$$f(x)=\left\{\begin{array}{ll} \frac{1-\cos k x}{x^2}, & \text { if } x \leq 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text { if } x>0 \end{array}\right. \text { is continuous at }$$ $$x=0$$, then the value of $$\mathrm{k}$$ is

$$\lim _\limits{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$$ equals

$$\matrix{ {f(x) = a{x^2} + bx + 1,} & {if} & {\left| {2x - 3} \right| \ge 2} \cr { = 3x + 2,} & {if} & {{1 \over 2} < x < {5 \over 2}} \cr } $$

is continuous on its domain, then $$a+b$$ has the value

If $$\lim _\limits{x \rightarrow 1} \frac{x^2-a x+b}{(x-1)}=5$$, then $$(a+b)$$ is equal to

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x}=$$

If $$\mathrm{f}(\mathrm{x})=\mathrm{x}, \quad$$ for $$\mathrm{x} \leq 0$$

$$=0,\quad$$ for $$x>0$$, then the function $$f(x)$$ at $$x=0$$ is

$$\lim _\limits{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=$$

If the function

$$\begin{array}{rlrl} f(x) & =3 a x+b, & & \text { for } x<1 \\ & =11, & & \text { for } x=1 \\ & =5 a x-2 b, & \text { for } x>1 \end{array}$$

is continuous at $$x=1$$. Then, the values of $$a$$ and $$b$$ are

$$\begin{aligned} & \text { If the function given by} \mathrm{f}(\mathrm{x}) \\ & =-2 \sin \mathrm{x} \quad-\pi \leq \mathrm{x}<-(\pi / 2) \\ & =a \sin x+b \quad-(\pi / 2)< x<(\pi / 2) \\ & =\cos x \quad(\pi / 2) \leq x \leq \pi \\ \end{aligned}$$

is continuous in $$[-\pi, \pi]$$, then the value of $$(3 a+2 b)^3$$ is

If $$f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$$, for $$x \neq \pi$$ is continuous at $$x=\pi$$, then the value of $$f(\pi)$$ is

$$\lim _\limits{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]=$$

Let

$$\begin{aligned} f(x) & =x+a \sqrt{2} \sin x & & , 0 \leq x<\frac{\pi}{4} \\ & =2 x \cot x+b & & \frac{\pi}{4} \leq x<\frac{\pi}{2} \\ & =a \cos 2 x-b \sin x & & \frac{\pi}{2} \leq x \leq \pi \end{aligned}$$

If $$\mathrm{f}(\mathrm{x})$$ is continuous for $$0 \leq \mathrm{x} \leq \pi$$, then

$$\lim _\limits{x \rightarrow 2}(x-1)^{ \frac{1}{3 x-6}}=$$

Let

$$f(x)\matrix{ { = |x| + 3,} & {if\,x \le - 3} \cr { = - 2x,} & {if\, - 3 < x < 3} \cr { = 6x - 2,} & {if\,x \ge 3} \cr } $$, then

$$\lim _\limits{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2}=$$

$$\begin{aligned} & \text { } f(x)=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \text {, if } 1 \leq x<0 \\ & =\frac{2 x+1}{x-2} \quad \text {, if } 0 \leq x \leq 1 \\ \end{aligned}$$

is continuous in the interval $$[-1,1]$$, then $$p=$$

If $$\lim _\limits{x \rightarrow 5} \frac{x^k-5^k}{x-5}=500$$, then the value of $$k$$, where $$k \in N$$ is

$$\begin{aligned} & \text { If the function } \mathrm{f}(\mathrm{x})=1+\sin \frac{\pi}{2}, \quad-\infty<\mathrm{x} \leq 1 \\ & =\mathrm{ax}+\mathrm{b}, \quad 1<\mathrm{x}<3 \\ & =6 \tan \frac{x \pi}{12}, \quad 3 \leq x<6 \\ \end{aligned}$$

is continuous in $$(-\infty, 6)$$, then the values of $$\mathrm{a}$$ and $$\mathrm{b}$$ are respectively.

$$\lim _\limits{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$$

If $$a=\lim _\limits{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$$ and $$b=\lim _\limits{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$$, then

If $$f(x) = {{{4^{x - \pi }} + {4^{x - \pi }} - 2} \over {{{(x - \pi )}^2}}}$$, for $$x \ne \pi $$, is continuous at $$x=\pi$$, then k =

$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + 5x - 7} - x} \right) = $$

If f(x) = |x|, for x $$\in$$ ($$-1,2$$), then f is discontinuous at (where [x] represents floor function)

If $f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$, for $x \neq \pi$ is continuous at $x=\pi$, then $f(\pi)=$

If $$f(x)=\frac{|x|}{x}$$, for $$x \neq 0$$ $$=1$$, for $$x=0$$, then tre function is

The function $$f(x)=\frac{x+1}{9 x+x^3}$$ is

The points of discontinuity of the function

$$\begin{aligned} f(x) & =\frac{1}{x-1}, \text { if } 0 \leq x \leq 2 \\ & =\frac{x+5}{x+3} \text { if } 2< x \leq 4 \end{aligned}$$

in its domain are

If $f(x)$ is continuous at $x=3$, where

$$\begin{aligned} f(x) & =a x+1, & \text { for } x \leq 3 \\ & =b x+3 & , \text { for } x>3 \text { then } \end{aligned}$$

$$\begin{aligned} & \text { If } f(x)=\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}, \quad x \neq 0 \\ & =k \text {, } \qquad x=0 \text { is continuous }\\ & x=0 \end{aligned}$$ Then $k=$

If the function $f(x)=\frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}, x \neq 0$

$$\qquad \qquad=16 \qquad x=0$$

is continuous at $x=0$, then $k=\ldots \ldots$

If $f(x)=[x]$, where $[x]$ is the greatest integer not greater than $x$, then $f^{\prime}\left(1^{+}\right)=$ ...........

If function

$$\begin{aligned} f(x) & =x-\frac{|x|}{x}, x<0 \\ & =x+\frac{|x|}{x}, x>0 \\ & =1, \quad x=0, \text { then } \end{aligned}$$

Which of the following function is not continuous at $x=0$ ?

If the function $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}$ $x \neq 0$ is continuous at $x=0$ then, $f(0)=\ldots \ldots$