Limits, Continuity and Differentiability

The domain of the derivative of the function $f(x)=\frac{x}{1+|x|}$ is

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

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,
\(\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=\)

\(\mathop {\lim }\limits_{x \to {\pi \over 4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=\)

\(\lim\limits_{x \rightarrow \infty}[x-\log (\cosh x)]=\)

\(\lim\limits_{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)=\)

If a real valued function $f(x)=\left\{\begin{array}{cl}e^{\frac{\sin a(x-[x])}{x-[x]}} & , \text { if } x<1 \\ b+1 & , \text { if } x=1 \text { is } \\ \frac{\left|x^2+x-2\right|}{x-1} & , \text { if } x>1\end{array}\right.$ continuous at ...

If a real valued function
$$ f(x)=\left\{\begin{array}{cc} \log (1+[x]), & x \geq 0 \\ \sin ^{-1}[x], & -1 \leq x<0 \\ k([x]+|x|), & x<-1 \end{array}\right. $$
is continuous at $x=-1$, then $k=$

\(\mathop {\lim }\limits_{x \to \infty } \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}}=\)

\(\mathop {\lim }\limits_{n \to \infty } \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\sin \frac{3 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=\)

\(\mathop {\lim }\limits_{x \to 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3}-\sqrt{\sin ^2 x-2 \tan x-x+3}}\)

\(\mathop {\lim }\limits_{x \to 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}=\)

$[x]$ represents the greatest integer function. If $\mathop {\lim }\limits_{x \to 0 + } \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$, then $k=$

If $f(x)=\left\{\begin{array}{cl}\frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x}, & \text { if } x>0 \\ 2, & \text { if } x=0 \\ \frac{\cos 4 x-\cos b x}{\tan ^2 x}, & \text { if } x<0\end{array}\right.$ is continuous at $x=0$, then $\s...

\(\mathop {\lim }\limits_{x \to \infty } \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}}=\)

The quadratic equation whose roots are $l=\lim\limits_{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\theta}\right)$ and $m=\lim\limits_{\theta \rightarrow 0}\left(\frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}...

If a function,
$$ f(x)=\left\{\begin{array}{cc} \frac{\sqrt[3]{1+a x^2+b x^3}-\sqrt[3]{1-a x^2-b x^3}}{x^2}, & x<0 \\ 5, & x=0 \\ \frac{\tan 3 x-\sin 3 x}{b x^3}, & x>0 \end{array}\right. $$
is continuous at $x=0$, then the geometric mean o...

\(\mathop {\lim }\limits_{x \to 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=\)

If a real valued function
$$ f(x)=\left\{\begin{array}{cc} (1+\sin x)^{\cos x}, & -\pi / 2 < x < 0 \\ a, & x=0 \\ \frac{e^{2 / x}+e^{3 / x}}{a e^{2 / x}+b e^{3 / x}}, & 0 < x < \pi / 2 \end{array}\right. $$
is continuous at $x=0$, then $a b...

\(\mathop {\lim }\limits_{x \to 0} \frac{(\operatorname{cosec} x-\cot x)\left(e^x-e^{-x}\right)}{\sqrt{3}-\sqrt{2+\cos x}}=\)

\(\mathop {\lim }\limits_{y \to 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}=\)

If the function $f(x)=\left\{\begin{array}{l}1+\cos x, x \leq 0 \\ a-x, 02\end{array}\right.$ everywhere, then $a^2+b^2=$

If $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 4 x}{1-\cos 2 x}=k$, then $\lim\limits_{x \rightarrow k} \frac{x^k-27}{x^{k+1}-81}=$

Let $f: R \rightarrow R$ be defined by
$$ f(x)=\left\{\begin{array}{cc} a-\frac{\sin [x-1]}{x-1}, & \text { if } x>1 \\ 1, & \text { if } x=1 \\ b-\left[\frac{\sin [x-1]-[x-1]}{([x-1])^3},\right. & \text { if } x<1 \end{array}\right. $$
whe...

\(\mathop {\lim }\limits_{n \to \infty } \frac{1}{n^3} \sum\limits_{k=1}^n k^2 x=\)

$[x]$ denotes the greater integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim\limits_{x \rightarrow 0}-\frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$, then $\sin \theta+\cos \theta=$

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

\(\mathop {\lim }\limits_{x \to - \infty } \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}=\)

If $\mathop {\lim }\limits_{x \to {a^ + }} f(x)=p, \mathop {\lim }\limits_{x \to {a^ - }} f(x)=m$ and $f(a)=k$, then which one of the following is true?

Match the functions in Column I with their properties in Column II. In the following [ $x$ ] denotes the greatest integer less than or equal to $x$.
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\(\mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt{2})-\sqrt{1+\cos x}}{\sqrt{15+\cos 2 x-4}}=\)

\(\mathop {\lim }\limits_{x \to \infty }\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{n+2 n}{n^2+4 n^2}\right]=\)

Consider the following functions
I. $f(x)= \begin{cases}\frac{1}{2}-x & , x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2 & , x \geq \frac{1}{2}\end{cases}$
II. $f(x)=|3 x-1|$
III. $f(x)=x|x|$
IV. $f(x)=|x|$
Then, on $[0,1]$ Lagrange's mean v...

\(\mathop {\lim }\limits_{x \to 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}=\)

If a real valued function
$$ f(x)=\left\{\begin{array}{cl} \frac{x^2+(a+3) x+(a+1)}{x+3} & , \text { when } x \neq-3 \\ -\frac{5}{2} & , \text { when } x=-3 \end{array}\right. $$
is continuous at $x=-3$, then $\lim _{x \rightarrow a}\left(x...

The function $f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0\end{array}\right.$

If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \text { is differentiable } \\ b x^3+1, & x > 2\end{array}\right.$ $\forall x \in R$, then $a-b$ is equal to

If $f(x)=\left\{\begin{array}{cl}\frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x \neq 0 \text { is } \\ K & x=0\end{array}\right.$ continuous at $x=0$, then $K$ is equal to

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$, then the number of integral values of ' $a$ ' such that $\lim \limits_{x \rightarrow a}\left(\left[x^2\right]-[x]^2\right)$ does not exist is

Let $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \\ a x, & 1 < x \leq 2\end{array}\right.$.If $\lim _{x \rightarrow 1} f(x)$ exists, then the sum of the cubes of the possible values of $a$ is

In each of the following options, a function and an interval are given. Choose the option containing the function and the interval for which Lagrange's mean value theorem is not applicable

\(\mathop {\lim }\limits_{n \to \infty }\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)=\)

The function $f(x)=|x-24|$ is

If $f(x)=\left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x=0$, then $f(0)=$

Let $f(x)=\left\{\begin{array}{cl}0, & x=0 \\ 2-x, & \text { for } 0 < x < 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 < x < 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.$
then which of the following is t...

\(\mathop {\lim }\limits_{x \to o} \left[\frac{1}{x}-\frac{1}{e^x-1}\right]=\)

If a real valued function $f(x)=\left\{\begin{array}{cl}\frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}, & \text { for } x \neq 3 \\ 1, & \text { for } x=3\end{array}\right.$ is continuous at $x=3$ and $l$ is a finite value, then $l-k$ is equal to

$\mathop {\lim }\limits_{x \to 0}\frac{\cos 2 x-\cos 3 x}{4 x-\cos 5 x}$ is equal to $\cos 4 x-\cos 5 x$

\(\mathop {\lim }\limits_{x \to \infty } \frac{[2 x-3]}{x} \text { is equal to }\)

If

\(\lim _{x \rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\left(P x^4-16\right)}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8}=1\)

then $P=$