$\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x} = $

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

$\lim _{x}$ ${\rightarrow 0} \frac{\tan \left(\left[-\pi^2\right] x^2\right)-x^2 \tan \left(\left[-\pi^2\right]\right)}{\sin ^2 x}$ equals

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

$\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = $

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