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Question Number 194891 by cortano12 last updated on 18/Jul/23

$$\underset{x\to 3^{+}}{\lim }\left(\frac{\sqrt{x}-\sqrt{x-3}-\sqrt{3}}{\sqrt{x^2-9}}\right)=?$$

Answered by MM42 last updated on 18/Jul/23

$${lim}_{x\to 3}(\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x-3}\times \sqrt{x+3}}-\frac{\sqrt{x-3}}{\sqrt{x-3}\times \sqrt{x+3}})$$$$={lim}_{x\to 3}(\frac{\sqrt{x-3}\times \sqrt{x-3}}{(\sqrt{x}+\sqrt{3})\times \sqrt{x-3}\times \sqrt{x+3}}-\frac{\sqrt{x-3}}{\sqrt{x-3}\times \sqrt{x+3}})$$$$={lim}_{x\to 3}(\frac{\sqrt{x-3}}{(\sqrt{x}+\sqrt{3})\times \sqrt{x+3}}-\frac{1}{\sqrt{x+3}})$$$$=-\frac{1}{\sqrt{3}}✓$$$$$$

Commented by cortano12 last updated on 18/Jul/23

$$=-\frac{1}{\sqrt{3}}\left(\times \right)$$$$=-\frac{1}{\sqrt{6}}\left(✓\right)$$

Commented by MM42 last updated on 18/Jul/23

$$ok$$$$itwasmymistake$$$$x\to 3\Rightarrow -\frac{1}{\sqrt{x+3}}=-\frac{1}{\sqrt{6}}$$

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