calculus - $lim_{xrightarrow 6} frac{sqrt{x+3}-3}{x-6}$ without L'Hôpital's rule

I'm trying to do the following limit

$$\lim_{x\rightarrow 6} \frac{\sqrt{x+3}-3}{x-6}$$

without using L'Hôpital's rule.

Anyone knows any neat tricks that can be used?

Answer

Multiply by conjugate of numerator:

$$\frac{\sqrt{x+3}-3}{x-6}\cdot\frac{\sqrt{x+3}+3}{\sqrt{x+3}+3}=\frac{x-6}{(x-6)(\sqrt{x+3}+3)}=\frac1{\sqrt{x+3}+3}\xrightarrow[x\to6]{}\frac16$$