Limits

The value of $limx\rightarrow{1}^{-}\frac{\sqrt{π}-\sqrt{4ta{n}^{-1}x}}{\sqrt{1-x}}$ is equal to

If $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2$, then

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

The value of $limx→0\frac{xcot(4x)}{ta{n}^{2}(3x)co{t}^{2}(6x)}$ is equal to

$\lim_{x \to 1} \left[\dfrac{x-2}{x^2-x} - \dfrac{1}{x^3 - 3x^2 + 2x}\right] = $

If $f(x)=\frac{\sqrt{\pi}-\sqrt{\operatorname{Cos}^{-1} x}}{\sqrt{1+x}}$ for $x>-1$ then $\lim _{x \rightarrow-1^{+}} f(x)=$

If $\lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2} = \lim\limits_{x \to 0} \dfrac{1 - \cos(2x)}{x \sin x}$, then the value of $k$ is $\ldots$

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

$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\left(\cos ^{-1} x\right)^2}=$

Define $\mathrm{f}(x)=\left\{\begin{array}{cl}\mathrm{b}-\mathrm{a} x & , \text { if } x 2\end{array}\right.$ and if $\lim _{x \rightarrow 2} \mathrm{f}(x)$ exists, then $\frac{\mathrm{a}}{\mathrm{b}}=$

If $\quad \mathbf{f}(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2} & \text { if } x 0\end{array}\right.$ is continuous at $x=0$ Then $a=$

$\lim _{x \rightarrow 0}\left(1^{\operatorname{cosec}^2 x}+2^{\operatorname{cosec}^2 x}+\ldots+n^{\operatorname{cosec}^2 x}\right)^{\sin ^2 x}$ is equal to