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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Sorry, that was to early. Deleted my answer... – frog Aug 05 '14 at 11:48
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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 05 '14 at 12:52
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1Actually 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 05 '14 at 13:38
4 Answers
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$$\eqalign{ \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}} &= \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{\sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{(\sqrt x - \sqrt2)(\sqrt x + \sqrt 2)}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}} + \frac{x - 2}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}}+\frac{\sqrt{x-2}}{\sqrt{x+2}(\sqrt x + \sqrt 2)},}$$
which is easy to find the limit of.
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@Hakim Thanks for the edit! Had no idea how to do that. Still on the learning curve for Latex. Much obliged. :) – Deepak Aug 05 '14 at 12:02
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We have a $0/0$ form. Apply L'Hospital once after solving we get the answer $1/2$.
Mark Fantini
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user159480
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I don't know why this answer is downvoted. It looks fine to me. – Steven Van Geluwe Aug 05 '14 at 13:12
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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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Deepak's answer is better, but you can also do as the followings :
$$\begin{align}\frac{(\sqrt{x-2}+\sqrt x)-\sqrt 2}{\sqrt{x^2-4}}\cdot \frac{(\sqrt{x-2}+\sqrt x)+\sqrt 2}{(\sqrt{x-2}+\sqrt x)+\sqrt 2}\end{align}$$ $$=\frac{x-2+2\sqrt{x(x-2)}+x-2}{\sqrt{(x-2)(x+2)}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$ $$=\frac{\color{red}{\sqrt{x-2}}\ (\sqrt{x-2}+2\sqrt x+\sqrt{x-2})}{\color{red}{\sqrt{x-2}}\sqrt{x+2}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$
mathlove
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