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Question Number 112637 by bemath last updated on 09/Sep/20

$$\left(1\right)\underset{x\to 0}{\lim }{\left(\frac{\sinh \mathrm{x}}{\mathrm{x}}\right)}^{\frac{1}{{\mathrm{x}}^2}}?$$$$\left(2\right)\underset{x\to 0}{\lim }\mathrm{x}\ln \left(\tan \mathrm{x}\right)?$$

Answered by john santu last updated on 09/Sep/20

$$\left(2\right)L=\underset{x\to 0}{\lim }x\ln \left(\tan x\right)$$$$L=\underset{x\to 0}{\lim }\frac{\ln \left(\tan x\right)}{\left(\frac{1}{x}\right)}$$$$L=\underset{x\to 0}{\lim }\frac{\frac{\sec {}^{2}x}{\tan x}}{-\frac{1}{x^2}}=-\underset{x\to 0}{\lim }\frac{x^2}{\sin x\cos x}$$$$L=-\underset{x\to 0}{\lim }\frac{x}{\cos x}=0$$

Answered by john santu last updated on 09/Sep/20

$$\left(1\right)L=\underset{x\to 0}{\lim }{\left(\frac{\sinh x}{x}\right)}^{\frac{1}{x^2}}$$$$\ln L=\underset{x\to 0}{\lim }\frac{1}{x^2}\ln \left(\frac{\sinh x}{x}\right)$$$$\ln L=\underset{x\to 0}{\lim }\frac{\ln (1+\frac{x^2}{6}+\frac{x^4}{120}+...)}{x^2}$$$$\ln L=\underset{x\to 0}{\lim }\left(\frac{\frac{d}{dx}\ln (1+\frac{x^2}{6}+\frac{x^4}{120}+...)}{2x}\right)$$$$\ln L=\underset{x\to 0}{\lim }\frac{1}{2x}.\frac{\frac{2x}{6}+\frac{4x^3}{120}+...}{1+\frac{x^2}{6}+\frac{x^4}{120}+...}$$$$\ln L=\underset{x\to 0}{\lim }\frac{\frac{1}{6}+\frac{2x^3}{120}+...}{1+\frac{x^2}{6}+\frac{x^4}{120}+...}$$$$\ln L=\frac{1}{6}\Rightarrow L=\sqrt[6]{e}.$$

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