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How to evaluate the value of this limit?

$$\lim_{x\to 2} \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}}$$

Actually I'm struck at algebraic part. Please guide..

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closed as off-topic by Antonio Vargas, apnorton, heropup, Semiclassical, Claude Leibovici Aug 10 '14 at 4:02

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If this question can be reworded to fit the rules in the help center, please edit the question.

Sorry, that was to early. Deleted my answer... – frog Aug 5 '14 at 11:48
What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. – user1729 Aug 5 '14 at 12:52
Actually I was struck at the 3rd step of the answer @Deepak told. Don't worry I complete my homework myself. There's no benefit of asking without trying at my fullest as in that case I'll get no benefits of that particular problem. Thanks for pointing.. :) – Adarsh Aug 5 '14 at 13:38

4 Answers 4

up vote 5 down vote accepted

$$\eqalign{ \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}} &= \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{\sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{(\sqrt x - \sqrt2)(\sqrt x + \sqrt 2)}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}} + \frac{x - 2}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}}+\frac{\sqrt{x-2}}{\sqrt{x+2}(\sqrt x + \sqrt 2)},}$$

which is easy to find the limit of.

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+1 Oh, it's better than mine! – mathlove Aug 5 '14 at 11:56
@Hakim Thanks for the edit! Had no idea how to do that. Still on the learning curve for Latex. Much obliged. :) – Deepak Aug 5 '14 at 12:02
@Deepak You're welcome, you will rapidly get used to it! ;-) – Hakim Aug 5 '14 at 12:02
Thanks a lot brother... got it now. :) – Adarsh Aug 5 '14 at 12:05
@Adarsh. You're most welcome brother. :) – Deepak Aug 5 '14 at 12:09

Deepak's answer is better, but you can also do as the followings :

$$\begin{align}\frac{(\sqrt{x-2}+\sqrt x)-\sqrt 2}{\sqrt{x^2-4}}\cdot \frac{(\sqrt{x-2}+\sqrt x)+\sqrt 2}{(\sqrt{x-2}+\sqrt x)+\sqrt 2}\end{align}$$ $$=\frac{x-2+2\sqrt{x(x-2)}+x-2}{\sqrt{(x-2)(x+2)}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$ $$=\frac{\color{red}{\sqrt{x-2}}\ (\sqrt{x-2}+2\sqrt x+\sqrt{x-2})}{\color{red}{\sqrt{x-2}}\sqrt{x+2}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$

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Thanks bro.. :) – Adarsh Aug 5 '14 at 12:05
@Adarsh: You are welcome! – mathlove Aug 5 '14 at 12:06

We have a $0/0$ form. Apply L'Hospital once after solving we get the answer $1/2$.

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I don't know why this answer is downvoted. It looks fine to me. – Steven Van Geluwe Aug 5 '14 at 13:12

A solution with a bit less algebra: $$\lim_{x \to 2}\frac{\sqrt{x-2}+\sqrt{x}-2}{\sqrt{x-2}\sqrt{x+2}}=\lim_{x \to 2}\frac{1+\frac{\sqrt{x}-2}{\sqrt{x-2}}}{\sqrt{x+2}}=\lim_{x \to 2}\frac{1}{\sqrt{x+2}} + \lim_{x \to 2}\frac{\sqrt{x}-2}{\sqrt{x^2-4}}=$$ For the computation of the right limit, you should use de l'Hôpital's rule (it seems to me the only appropriate way): $$\frac{1}{2}+\lim_{x \to 2}\frac{\sqrt{x^2-4}}{2x\sqrt x}=\frac{1}{2}+0=\frac{1}{2}$$

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