Evaluate the limit: $\lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)$

$\lim _{x \rightarrow 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}$ is equal to

$\lim _{x \rightarrow 0} \frac{\cos ax - \cos bx}{x^{2}}$ is equal to

$\mathop {\lim }\limits_{h \to 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)} \right]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}} = $