Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$,where $a, b \neq 0$.

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx}$

$\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)}} = $

$\mathop {\lim }\limits_{\theta \to 0} \frac{{5\theta \cos \theta - 2\sin \theta }}{{3\theta + \tan \theta }} = $

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to $.......$

The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$ is