$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos 6x}{x} = $

The value of $ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $ is

The value of the limit $\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ is

$\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\} =$

$\lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$ is equal to :

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