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Question Number 131383 by EDWIN88 last updated on 04/Feb/21

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x^2}{x-1}\right)}^{\tan \left(\frac{1}{\sqrt{x}}\right)}=?$$

Answered by liberty last updated on 04/Feb/21

$$\mathrm{L}=\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{{\mathrm{x}}^2}{\mathrm{x}-1}\right)}^{\tan \left(\frac{1}{\sqrt{\mathrm{x}}}\right)}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }\left(\tan \frac{1}{\sqrt{\mathrm{x}}}\right)\ln \left(\frac{{\mathrm{x}}^2}{\mathrm{x}-1}\right)}={\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{2\ln \mathrm{x}-\ln (\mathrm{x}-1)}{\cot \left(\frac{1}{\sqrt{\mathrm{x}}}\right)}\right)}$$$$\mathrm{L}={\mathrm{e}}^{\underset{\mathrm{t}\to 0}{\lim }\left(\frac{4\ln \mathrm{t}-\ln ({\mathrm{t}}^2-1)}{\cot \left(\frac{1}{\mathrm{t}}\right)}\right)}=\mathrm{e}{}^{\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\left(\frac{2({\mathrm{t}}^2-1)}{\mathrm{t}({\mathrm{t}}^2-1)\left(\frac{1}{{\mathrm{t}}^2}\right)\mathrm{cosec}{}^2\left(\frac{1}{\mathrm{t}}\right)}\right)}$$$$\mathrm{L}={\mathrm{e}}^{\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }{\left(\frac{\sin \left(\frac{1}{\mathrm{t}}\right)}{\frac{1}{\mathrm{t}}}\right)}^2.\left(\frac{{2\mathrm{t}}^2-4}{{\mathrm{t}}^3-\mathrm{t}}\right)}={\mathrm{e}}^0=1$$

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