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

Question

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.

Answer 1

$$\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt 2}$$ $$=\lim_{x\to 4}\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)}{(\sqrt{x-2}-\sqrt 2)(\sqrt{x-2}+\sqrt 2)}\cdot \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$ $$=\lim_{x\to 4}\frac{2x+1-9}{x-2-2}\cdot \lim_{x\to 4} \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$ $$=\lim_{x\to 4}\frac{2x-8}{x-4}\cdot \frac{\sqrt{2}+\sqrt 2}{3+3}$$ $$=\frac{\sqrt 2}{3}\lim_{x\to 4}\frac{2(x-4)}{x-4}$$ $$=\frac{\sqrt 2}{3}\lim_{x\to 4} 2=\color{red}{\frac{2\sqrt 2}{3}}$$

Answer 2

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.