Question

How to do question 336....please solve fast
بما​

Answer 1

Answer:

Mark as brainliest if it helps

Answer 2

Question :-

Solve this limit .

$\leadsto \: \lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \sin \big(x - \frac{ \pi}{6} \big)}{ \frac{ \sqrt{3} }{2} - \cos x }$

Answer :-

$\implies \: \frac{ 3\sqrt{3} +\pi}{2\pi} \:$

Explanation :-

We have -

$\leadsto \: \lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \sin \big(x - \frac{ \pi}{6} \big)}{ \frac{ \sqrt{3} }{2} - \cos x } \\ \:$

apply LH rule ( L- hospital )

LH rule :-

→ Here we do differentiation of denominator and nominator.

Hence ,

$\implies \sf{\lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \frac{d}{dx} \sin \big(x - \frac{ \pi}{6} \big)}{ \frac{d}{dx} \big( \frac{ \sqrt{3} }{2} - \cos x \big)} }\\ \\ \because \boxed{ \sf{ \frac{d}{dx} \sin x = \cos x \: }} \\ and \: \\ \: \: \: \: \sf{ \boxed{ \sf{\frac{d}{dx} \cos x = - \sin x}}} \\ \\ \to \: \lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \cos \big(x - \frac{ \pi}{6} \big)}{ \sin x } \\ \:$

and , we know that

→ Cos (A+B) = Cos A .Cos B + Sin A.Sin B

Hence ,

$\to \: \lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \cos x. \cos \frac{ \pi}{6} + \sin x \sin \frac{ \pi}{6} }{ \sin x } \\ \: \: \\ \to \lim_{x\to \: \frac{ \pi}{6} } \bigg( \frac{ \sqrt{3} }{2\tan x} + \frac{1}{2} \bigg) \\ \\ \to \: \lim_{x\to \: \frac{ \pi}{6} } \bigg( \frac{ \sqrt{3} }{2} \frac{x }{x\tan x} + \frac{1}{2} \bigg) \\ \\ \to \: \lim_{x\to \: \frac{ \pi}{6} } \: \frac{ \sqrt{3} }{2x} + \frac{1}{2} \\ \\ \bf{ taking \: limit \:} \\ \\ \to \: \frac{ \sqrt{3} }{2} \times \frac{6}{\pi} + \frac{1}{2} \\ \\ \to \: \frac{ 3\sqrt{3} }{\pi} + \frac{1}{2} = \frac{ 3\sqrt{3} +\pi}{2\pi}$