If $f(x) = \begin{cases} x & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$,then $\mathop {\lim }\limits_{x \to 0} f(x) = $

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{(x+1)^{5}-1}{x}$

If $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ and $\lim _{x \rightarrow 0^{-}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=l$,then which of the following is true?

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x} = $

$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}} = $