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

I am wondering how to find $\lim\limits_{x~\to~4}\dfrac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}.$

I have found out that it equals $\dfrac{0}{0}$ if the equation is not simplified.

I have tried multiplying by the conjugate of the numerator (and the denominator in a separate attempt), but I have not found any other solution other than $\dfrac{0}{0}.$

Does anyone know how to find this limit? If so, could they please show the steps they used to get to the answer?

All help is appreciated.

• @HarishChandraRajpoot I'm pretty sure MSE doesn't need many of the tags you've created over the past several days (and I've removed most I've seen). Why not use, for example, radicals which already exists? – pjs36 Oct 2 '15 at 23:59

$$\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt 2}$$ $$=\lim_{x\to 4}\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)}{(\sqrt{x-2}-\sqrt 2)(\sqrt{x-2}+\sqrt 2)}\cdot \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$ $$=\lim_{x\to 4}\frac{2x+1-9}{x-2-2}\cdot \lim_{x\to 4} \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$ $$=\lim_{x\to 4}\frac{2x-8}{x-4}\cdot \frac{\sqrt{2}+\sqrt 2}{3+3}$$ $$=\frac{\sqrt 2}{3}\lim_{x\to 4}\frac{2(x-4)}{x-4}$$ $$=\frac{\sqrt 2}{3}\lim_{x\to 4} 2=\color{red}{\frac{2\sqrt 2}{3}}$$
You need to multiply by the conjugates of numerator and denominator at the same time. Then in the fraction you will have $(x-4)$ in both numerator and denominator, and they will cancel out.