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Question Number 122235 by benjo_mathlover last updated on 15/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x}{x+2}\right)}^x?$$

Answered by liberty last updated on 15/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{\mathrm{x}}{\mathrm{x}+2}\right)}^{\mathrm{x}}={\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }(\frac{\mathrm{x}}{\mathrm{x}+2}-1).\mathrm{x}}$$$$={\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{-2\mathrm{x}}{\mathrm{x}+2}\right)}={\mathrm{e}}^{-2}=\frac{1}{{\mathrm{e}}^2}.$$

Answered by Dwaipayan Shikari last updated on 15/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{2}{x+2})}^x={(1-\frac{2}{x+2})}^{\frac{-2x(x+2)}{-2(x+2)}}=e^{-\frac{2x}{(x+2)}}=e^{-2}=\frac{1}{e^2}$$

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