Evaluation of limit 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..
 A: $$\eqalign{ \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}} &= \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{\sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{(\sqrt x - \sqrt2)(\sqrt x + \sqrt 2)}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}} + \frac{x - 2}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}}+\frac{\sqrt{x-2}}{\sqrt{x+2}(\sqrt x + \sqrt 2)},}$$
which is easy to find the limit of.
A: We have a $0/0$ form. Apply L'Hospital once after solving we get the answer $1/2$.
A: 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):
$$\frac{1}{2}+\lim_{x \to 2}\frac{\sqrt{x^2-4}}{2x\sqrt x}=\frac{1}{2}+0=\frac{1}{2}$$
A: Deepak's answer is better, but you can also do as the followings :
$$\begin{align}\frac{(\sqrt{x-2}+\sqrt x)-\sqrt 2}{\sqrt{x^2-4}}\cdot \frac{(\sqrt{x-2}+\sqrt x)+\sqrt 2}{(\sqrt{x-2}+\sqrt x)+\sqrt 2}\end{align}$$
$$=\frac{x-2+2\sqrt{x(x-2)}+x-2}{\sqrt{(x-2)(x+2)}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$
$$=\frac{\color{red}{\sqrt{x-2}}\ (\sqrt{x-2}+2\sqrt x+\sqrt{x-2})}{\color{red}{\sqrt{x-2}}\sqrt{x+2}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$
