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Question Number 121460 by bramlexs22 last updated on 08/Nov/20

$$\underset{x\to 0^{+}}{\lim }{\left(\cot \mathrm{x}\right)}^{\frac{1}{\ln \mathrm{x}}}?$$

Answered by Dwaipayan Shikari last updated on 08/Nov/20

$$\underset{x\to 0}{\lim }{\left(cotx\right)}^{\frac{1}{logx}}=y$$$$\frac{1}{logx}log\left(\frac{cosx}{sinx}\right)=logy(cosx\to 1)$$$$\Rightarrow -\frac{log\left(sinx\right)}{logx}=logy$$$$=-1=logy\Rightarrow y=\frac{1}{e}(sinx\to x)$$

Answered by liberty last updated on 08/Nov/20

$$\mathrm{L}=\underset{x\to 0^{+}}{\lim }{\left(\cot \mathrm{x}\right)}^{\frac{1}{\ln \mathrm{x}}}$$$$\ln \mathrm{L}=\underset{x\to 0^{+}}{\lim }\frac{\ln \cot \mathrm{x}}{\ln \mathrm{x}}=\underset{x\to 0^{+}}{\lim }\frac{-\mathrm{cosec}{}^2\mathrm{x}}{\frac{1}{\mathrm{x}}.\cot \mathrm{x}}$$$$\ln \mathrm{L}=-\underset{x\to 0^{+}}{\lim }\frac{\mathrm{xsin}\mathrm{x}}{\sin {}^2\mathrm{x}.\cos \mathrm{x}}=-1$$$$\mathrm{L}={\mathrm{e}}^{-1}=\frac{1}{\mathrm{e}}.$$

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