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Question Number 169315 by Mastermind last updated on 28/Apr/22

$${lim}_{x\to \mathrm{\infty }}\left(\frac{1+\sqrt{x+2}}{1-\sqrt{x+2}}\right)$$$$$$$$Mastermind$$

Commented by infinityaction last updated on 28/Apr/22

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

Answered by alephzero last updated on 28/Apr/22

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1+\sqrt{x+2}}{1-\sqrt{x+2}}=$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1-\sqrt{x+2}}+\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x+2}}{1-\sqrt{x+2}}=$$$$=0+\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x+2}}{1-\sqrt{x+2}}$$$$\sqrt{x+2}\sim \sqrt{x+2}-1asx\to \mathrm{\infty }$$$$\Rightarrow \sqrt{x+2}\sim -(1-\sqrt{x+2})$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x+2}}{1-\sqrt{x+2}}=-1$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x+2}}{1-\sqrt{x+2}}=-1$$

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