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Question Number 143561 by ZiYangLee last updated on 15/Jun/21

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

Answered by mr W last updated on 15/Jun/21

$$a=\underset{x\to \mathrm{\infty }}{\lim }{\left[{(1-\frac{1}{x})}^2\right]}^x$$$$=\underset{x\to \mathrm{\infty }}{\lim }{\left[{(1-\frac{1}{x})}^{-x}\right]}^{-2}=e^{-2}=\frac{1}{e^2}$$$$$$$$b=\underset{n\to \mathrm{\infty }}{\lim }\frac{\sqrt{n+1}}{\sqrt{n}+\sqrt{n-1}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{\sqrt{1+\frac{1}{n}}}{1+\sqrt{1-\frac{1}{n}}}=\frac{1}{2}$$

Commented by ZiYangLee last updated on 15/Jun/21

$$wow...thankssirW!$$

Answered by Mathspace last updated on 15/Jun/21

$$f\left(x\right)={(1-\frac{2}{x}+\frac{1}{x^2})}^x\Rightarrow$$$$f\left(x\right)=e^{xlog(1-\frac{2}{x}+\frac{1}{x^2})}$$$$\sim e^{x(\frac{1}{x^2}-\frac{2}{x})}=e^{\frac{1}{x}-2}\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=e^{-2}=\frac{1}{e^2}$$$$$$

Answered by bobhans last updated on 15/Jun/21

$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x^2+x}-\sqrt{x^2-1}=\frac{1-0}{2\sqrt{1}}=\frac{1}{2}$$

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