$\mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\tan^{-1} 2x} \right) = $

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is equal to

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}\right]$

If $a > 0, b > 0$,then $\lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n =$

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$