# How do I solve this indeterminate limit without the L'hospital rule?

I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.

$$\lim\limits_{x\to 4} \frac{ \sqrt{2x+1}-3 }{ \sqrt{x-2}-\sqrt{2} }$$

By the way, the answer is supposed to be $\frac{2\sqrt{2}}{3}$.

• I'm surprised you've even heard of L'Hopital's rule if you haven't studied derivatives yet. – David H Apr 6 '14 at 0:44
• Rationalisation works if you do it for both numerator and denominator; then cancel the common terms. – Graham Kemp Apr 6 '14 at 1:01

$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} \\ = \lim_{x\to 4} \frac{2(x-4)(\sqrt{x-2}+\sqrt{2})}{(x-4)(\sqrt{2x+1}+3)} \\ = \lim_{x\to 4} \frac{2(\sqrt{x-2}+\sqrt{2})}{(\sqrt{2x+1}+3)} \\ = \frac{2\sqrt 2}{3}$$
Multiply top and bottom by $\sqrt{2x+1}+3$, and also by $\sqrt{x-2}+2$.