# finding the limit of an expression with roots

I need to find the limit of the expression: $$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}$$

It is given that $b>2$. In the class we were encouraged to use the method of multiplying both the numerator and the denominator with the numerator when the limit can't be found simply by using limit arithmetic. Hence: $$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2} =\\ =\lim_{x\to b}\frac{(\sqrt{b-2} - \sqrt{x-2})(\sqrt{b-2} - \sqrt{x-2})}{(x^2-b^2)(\sqrt{b-2} - \sqrt{x-2})} = \\ = \lim_{x\to b}\frac{b+x-4 - 2\sqrt{b+x-4}}{(x-b)(x+b)(\sqrt{b-2} - \sqrt{x-2})}$$

I don't see though how from this point this can be simplified. We could factor out the $\sqrt{b+x-4}$ in the numerator but this doesn't really help. Wolfram says the answer is $\frac{1}{4b\sqrt{b-2}}$

Multiply by $\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$ $$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$$ $$\lim_{x\to b}\frac{b-x}{(x-b)(x+b)(\sqrt{b-2} + \sqrt{x-2})}=\frac{-1}{(b+b)(\sqrt{b-2}+\sqrt{b-2})}=\frac{-1}{4b\sqrt{b-2}}$$
• Thank you! But in this transition $lim_{x\to b} \frac{-1}{(x+b)(\sqrt{b-2}+\sqrt{x-2)}}$ how do we get to $4b\sqrt{b-2}$ in the denominator?
• When we have $\frac{b-x}{(x-b)(x+b)}$, doesn't it become $\frac{-1}{x+b}$?
• @Yos yes and then let $x\rightarrow b$ Nov 19, 2016 at 11:41