# Evaluate limit without L'hopital

$$\lim_{x\to2}\dfrac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}$$

Please help me evaluate this limit. I have tried rationalising it but it just can't work, I keep ending up with $0$ at the denominator... Thanks!

• Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Sep 18 '15 at 13:21

$$\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}\times \frac{\sqrt{x^2+1}+\sqrt{2x+1}}{\sqrt{x^2+1}+\sqrt{2x+1}}\times \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^3-x^2}+\sqrt{x+2}}$$
So we get $$\displaystyle \lim_{x\rightarrow 2}\left[\frac{x^2-2x}{x^3-x^2-x-2}\times \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^2+1}+\sqrt{2x+1}}\right]$$
So we get $$\displaystyle \lim_{x\rightarrow 2}\frac{x(x-2)}{(x-2)\cdot (x^2+x+1)}\times \lim_{x\rightarrow 2} \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^2+1}+\sqrt{2x+1}}$$
So we get $$\displaystyle = \frac{2}{7}\times \frac{4}{2\sqrt{5}} = \frac{4}{7\sqrt{5}}$$