$\mathop {\lim }\limits_{x \to 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$ ની કિંમત શોધો.

આપેલ લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx}$

$\lim _{x \rightarrow \pi} \frac{1-\sin (x/2)}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)} =$

લક્ષ $\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}} = $

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