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

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\mathop {\lim }\limits_{x \to 1} (1 - x + [x - 1] + [1 - x])$ is

$\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2} = $

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

$\lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} = $

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then: