Limits Continuity And Differentiability - Calculus - Mathematics Previous Year Questions

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

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

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

$$\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 }$$

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

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

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

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

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

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

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 { 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 \(x=\frac{\pi}{2}\), then \(\lambda=\)

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

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