$\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ is

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

The value of $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to

$\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$

$[x]$ denotes the greatest integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$,then $\sin \theta+\cos \theta=$

$\mathop {\lim }\limits_{x \to 0} \frac{x}{|x| + {x^2}} = $