Limits, Continuity and Differentiability

The value of $\mathop {\lim }\limits_{x \to 3}\left[\frac{1}{x-3}+\frac{9 x}{27-x^3}\right]$ is:

If $f(x)=\left\{\begin{array}{l}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sin x} \\ \boldsymbol{k}, x=0\end{array}, x \neq 0\right.$ is continuous at $x=0$, then $\boldsymbol{k}=$

The function $y=||x|-1|$ is differentiable for all values of ' $x$ ' except

\(\mathop {\lim }\limits_{x \to 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \text { is equal to: }\)

The function $\boldsymbol{f}(\boldsymbol{x})=|\boldsymbol{x}|+|\boldsymbol{x}-\mathbf{1}|$ is:

If $\mathop {\lim }\limits_{x \to 0}\left(\frac{p \sin 2 x+1-\cos 2 x}{x+\tan x}\right)=1$ then the value of ' $p$ ' is

\(\mathop {\lim }\limits_{x \to {\pi \over 2}}\left(\frac{1-\sin x}{\cos x}\right) \text { is equal to }\)

If $f(x)=\left\{\begin{array}{ll}\frac{1-x^m}{1-x} & \text { if } x \neq 1 \\ 2 m-1 & \text { if } x=1\end{array}\right.$ and the function is discontinuous at $x=1$, then

The value of $\lim _\limits{x \rightarrow 0} \frac{(1-x)^n-1}{x}=$

Find the value of $\lim\limits_{h \rightarrow 0} \frac{(a+h)^2 \sin (a+h)-a^2 \sin a}{h}$

If $\lim\limits_{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim\limits_{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$, then the value of K is :

Evaluate: $\lim _\limits{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}$

The relationship between a and b for the continuous function
$f(x)=\left\{\begin{array}{ll}a x+1, & \text { if } x \leq 3 \\ b x+3, & \text { if } x>3\end{array}\right.$ at $x=3$ is

The function $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ is discontinuous at

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

$\lim _\limits{\theta \rightarrow \frac{\pi}{2}} \frac{1-\sin \theta}{\left(\frac{\pi}{2}-\theta\right) \cos \theta}$ is equal to :

\(\lim _\limits{x \rightarrow 0}\left(\frac{\sin a x}{\sin b x}\right)^k \text { equals }\)

$$ \text { If } f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{(\pi-2 x)^2} & , \quad \text { if } x \neq \frac{\pi}{2} \\ \lambda, & \text { if } x=\frac{\pi}{2} \end{array}\right. $$
Then \(f(x)\) will be continues function at $$x=\frac{\p...

\(\lim _\limits{x \rightarrow 0} \frac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a(a+x)}}\) equals to

$$ \text { If } f(x)=\left\{\begin{array}{cc} x & , \quad 0 \leq x \leq 1 \\ 2 x-1 & , \quad x>1 \end{array}\right. \text { then } $$

\(\text { The value of } \lim _\limits{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}=\)

Let \(\alpha\) and \(\beta\) be the distinct roots of \(a x^2+b x+c=0\), then \(\lim _\limits{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}\) is equal to

\(\text { The number of points of discontinuity of the rational function } f(x)=\frac{x^2-3 x+2}{4 x-x^3}\)

\(\text { The value of } \lim _\limits{x \rightarrow 0} \frac{\sin (a+x)-\sin (a-x)}{x} \text { is }\)

\(\lim _\limits{x \rightarrow 0} \frac{a^x-b^x}{c^x-d^x}=\)

\(\lim _\limits{x \rightarrow 0} \frac{a^x-b^x}{x} \text { is equal to }\)

$$
\text { The function defined by } f(x)=\left\{\begin{array}{cc}
\frac{\sin x}{x}+\cos x & x>0 \\
-5 k & x=0 \\
\frac{4(1-\sqrt{1-x})}{x} & x<0
\end{array} \quad \text { is continous at } x=0, \quad \text { then } k\right. \text { equals ...

The value of \(\lim _\limits{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}\) is

If \(f(x) = \left\{ {\matrix{ {2\sin x} & ; & { - \pi \le x \le {{ - \pi } \over 2}} \cr {a\sin x + b} & ; & { - {\pi \over 2} < x < {\pi \over 2}} \cr {\cos x} & ; & {{\pi \over 2} \le x \le \pi } \cr } } \right.\) and...

The value of \(\lim _\limits{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{2 x}\) is equal to

If \(\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}\), then

If the derivative of the function \(f(x) = \left\{ {\matrix{ {b{x^2} + ax + 4;} & {x \ge - 1} \cr {a{x^2} + b;} & {x < - 1} \cr } } \right.\) is everywhere continuous, then

If \(\mathop {\lim }\limits_{x \to 0} {{(1 + {a^3}) + 8{e^{1/x}}} \over {1 + (1 - {b^3}){e^{1/x}}}} = 2\), then

The value of \(\mathop {\lim }\limits_{x \to 0} \left( {{{{a^x} + {b^x} + {c^x}} \over 3}} \right),(a,b,c > 0)\) is

If \(f(x) = \left\{ {\matrix{ {ax + 3,} & {x \le 2} \cr {{a^2}x - 1} & {x > 2} \cr } } \right.\), then the values of a for which f is continuous for all x are

If \(L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}},a > 0\). If L is finite, then

If the function \(f(x) = \left\{ {\matrix{ {{{1 - \cos x} \over {{x^2}}},} & {\mathrm{for}\,x \ne 0} \cr {k,} & {\mathrm{for}\,x = 0} \cr } } \right.\) is continuous at x = 0, then the value of k is

\(\mathop {\lim }\limits_{x \to 1} {{\tan ({x^2} - 1)} \over {x - 1}}\) is equal to