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

If $$\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{x-3}{|x-3|}+\mathrm{a} & , \quad x 3\end{array}\right.$$ Is continuous at $$x=3$$, then the value of ...

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

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...

If $$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}, & -1 \leq x ...

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 $$...

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 $$x=0$$, t...

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

The value of $$\lim _\limits{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$$ 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} \si...

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

$$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{arra...

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

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

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

$$\begin{aligned} & \text { If the function } \mathrm{f}(\mathrm{x})=1+\sin \frac{\pi}{2}, \quad-\infty is continuous in $$(-\infty, 6)$$, then the va...

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

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}...

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 =

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

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