mathop {lim }limits_{x to pi /6} left[ {frac{ | vedclass

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Question

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o \pi /6} \frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}$.સ્વરૂપ $0/0$ હોવાથી,આપણે અંશ અને છેદનું $x$ ની સ

Vedclass · Accepted Answer

$1/\sqrt 3 $

Vedclass · Answer

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o \pi /6} \frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}$.સ્વરૂપ $0/0$ હોવાથી,આપણે અંશ અને છેદનું $x$ ની સાપેક્ષમાં વિકલન કરીને $L-Hospital$ નો નિયમ વાપરીએ:અંશનું વિકલન: $\frac{d}{dx}(3\sin x - \sqrt 3 \cos x) = 3\cos x + \sqrt 3 \sin x$.છેદનું વિકલન: $\frac{d}{dx}(6x - \pi) = 6$.$x 	o \pi /6$ લક્ષ લેતા:$\frac{3\cos(\pi /6) + \sqrt 3 \sin(\pi /6)}{6} = \frac{3(\sqrt 3 /2) + \sqrt 3 (1/2)}{6} = \frac{(3\sqrt 3 + \sqrt 3)/2}{6} = \frac{4\sqrt 3}{12} = \frac{\sqrt 3}{3} = \frac{1}{\sqrt 3}$.

$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $

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