Limits, Continuity and Differentiability

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

The number of discontinuities of the greatest integer function \(\mathrm{f}(x)=[x], x \in\left(-\frac{7}{2}, 100\right)\)

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

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

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

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

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} ...

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

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.$...

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

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 \(f(x)\) 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{arra...

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

\(\lim _\limits{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2}\) 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
...

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...

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

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 ...

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

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

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{...

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

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} \...

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

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

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

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

\(\matrix{ {f(x) = a{x^2} + bx + 1,} & {if} & {\left| {2x - 3} \right| \ge 2} \cr { = 3x + 2,} & {if} & {{1 \over 2} < x < {5 \over 2}} \cr }\)
is continuous on its domain, then \(a+b\) has the value

If 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

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

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

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

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

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

$$\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{ali...

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

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

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})...

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

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

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

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

$$\begin{aligned} & \text { } f(x)=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \text {, if } 1 \leq x<0 \\ & =\frac{2 x+1}{x-2} \quad \text {, if } 0 \leq x \leq 1 \\ \end{aligned}$$
is continuous in the interval \([-1,1]\), then \(p=\)

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

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

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 =

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

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