$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}} = $

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

$\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin 3\theta - \sin \theta }}{{\sin \theta }} = $

$\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}$ 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)}} = $