lim _{x rightarrow frac{pi}{2}} frac{1+cos 2 | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{3 \log 3}$

Vedclass · Answer

(C) આપેલ લક્ષ: $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)}$$1+\cos 2x = 2\cos^2 x$ અને $\cot 3x = \frac{\cos 3x}{\sin 3x}$ નો ઉપયોગ કરતા:$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^2 x \sin 3x}{\cos 3 x \left(3^{\sin 2 x}-1\right)}$ધારો કે $x = \frac{\pi}{2} + h$,જેમ $x \rightarrow \frac{\pi}{2}$,તેમ $h \rightarrow 0$. તેથી $\cos x = -\sin h$,$\cos 3x = \sin 3h$,$\sin 3x = -\cos 3h$,અને $\sin 2x = -\sin 2h$.આ કિંમતો મૂકતા:$L = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (3^{-\sin 2h} - 1)} = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (1 - 3^{-\sin 2h})}$$\lim_{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$ અને $\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$ નો ઉપયોગ કરતા:$L = \frac{1}{3 \ln 3}$.

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

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