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Question Number 119902 by benjo_mathlover last updated on 28/Oct/20

$$\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{\sin {}^{2}x}}?$$

Answered by benjo_mathlover last updated on 28/Oct/20

Answered by Olaf last updated on 28/Oct/20

$$\frac{1}{{\sin }^{2}x}\ln \left(\cos x\right)\underset{0}{\sim }\frac{1}{x^2}\ln (1-\frac{x^2}{2})\underset{0}{\sim }\frac{1}{x^2}\times (-\frac{x^2}{2})=-\frac{1}{2}$$$${\left(\cos x\right)}^{\frac{1}{{\sin }^{2}x}}\underset{0}{\sim }{e}^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$$

Answered by 1549442205PVT last updated on 28/Oct/20

$$\mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{\sin {}^{2}x}}$$$$\text{Missing left or extra right}$$$$=\mathrm{L}\underset{\mathrm{x}\to 0}{\mathrm{im}}\left(\frac{-1}{{2\cos }^2\mathrm{x}}\right)=\frac{-1}{2}$$$$\Rightarrow \mathrm{L}={\mathrm{e}}^{\frac{-1}{2}}=\frac{1}{\sqrt{\mathrm{e}}}$$

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