$\lim _{x \rightarrow 0} \frac{x \tan 4x - 2x \tan 2x}{(1 - \cos 4x)^2} = $

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

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)}=$

$\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)}} = $

सीमा का मूल्यांकन करें: $\lim _{x \rightarrow 0} \frac{\tan ^2(\pi \sec ^4 x)}{\pi^2 x^4}$

यदि $f(x) = -(\sin^2 x + \cos^5 x)$ है,तो $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$ का मान ज्ञात कीजिए।