lim_{h rightarrow 0} 2 left{ frac{sqrt{3} sin | vedclass

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Question

(A) ધારો કે $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \s

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) ધારો કે $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$.$\sin(A+B) = \sin A \cos B + \cos A \sin B$ અને $\cos(A+B) = \cos A \cos B - \sin A \sin B$ ના વિસ્તરણનો ઉપયોગ કરતા:અંશ $= \sqrt{3} (\sin \frac{\pi}{6} \cos h + \cos \frac{\pi}{6} \sin h) - (\cos \frac{\pi}{6} \cos h - \sin \frac{\pi}{6} \sin h)$$= \sqrt{3} (\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h) - (\frac{\sqrt{3}}{2} \cos h - \frac{1}{2} \sin h)$$= \frac{\sqrt{3}}{2} \cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2} \cos h + \frac{1}{2} \sin h = 2 \sin h$.લિમિટમાં કિંમત મૂકતા:$L = \lim_{h \rightarrow 0} 2 \left( \frac{2 \sin h}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right) = \frac{4}{\sqrt{3}} \lim_{h \rightarrow 0} \left( \frac{\sin h}{h} \right) \cdot \lim_{h \rightarrow 0} \left( \frac{1}{\sqrt{3} \cos h - \sin h} \right)$.$\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$ અને $\lim_{h \rightarrow 0} (\sqrt{3} \cos h - \sin h) = \sqrt{3}(1) - 0 = \sqrt{3}$ હોવાથી:$L = \frac{4}{\sqrt{3}} \cdot 1 \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}$.

$\lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$ ની કિંમત શોધો.

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