Limits, Continuity and Differentiability · Mathematics · COMEDK
MCQ (Single Correct Answer)
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
$$ \text { The value of } \lim _\limits{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}= $$
$$ \text { If } f(x)=\left\{\begin{array}{cc} x & , \quad 0 \leq x \leq 1 \\ 2 x-1 & , \quad x>1 \end{array}\right. \text { then } $$
$$ \text { If } f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{(\pi-2 x)^2} & , \quad \text { if } x \neq \frac{\pi}{2} \\ \lambda, & \text { if } x=\frac{\pi}{2} \end{array}\right. $$
Then $$f(x)$$ will be continues function at $$x=\frac{\pi}{2}$$, then $$\lambda=$$
$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a(a+x)}}$$ equals to
$$ \lim _\limits{x \rightarrow 0}\left(\frac{\sin a x}{\sin b x}\right)^k \text { equals } $$
The value of $$\lim _\limits{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{2 x}$$ is equal to
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 it is continuous on $$[-\pi, \pi]$$, then
The value of $$\lim _\limits{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}$$ is
$$ \lim _\limits{x \rightarrow 0} \frac{a^x-b^x}{x} \text { is equal to } $$
$$ \text { The function defined by } f(x)=\left\{\begin{array}{cc} \frac{\sin x}{x}+\cos x & x>0 \\ -5 k & x=0 \\ \frac{4(1-\sqrt{1-x})}{x} & x<0 \end{array} \quad \text { is continous at } x=0, \quad \text { then } k\right. \text { equals } $$
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
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
$$\mathop {\lim }\limits_{x \to 1} {{\tan ({x^2} - 1)} \over {x - 1}}$$ is equal to
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