$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$

$\lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{(2+\sqrt{2})^n - (2-\sqrt{2})^n} \right] =$

Evaluate the given limit: $\mathop {\lim }\limits_{x \to \pi } \left(x-\frac{22}{7}\right)$

The value of $\lim_{n \to \infty} \frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}$ is

Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If $L$ and $R$ respectively denote the left-hand limit and the right-hand limit of $f(x)$ at $x=0$,then $\frac{32}{\pi^2}\left(L^2+R^2\right)$ is equal to: