Calculate $\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}$ After I just learned there are ways to cancel out $0$s in the divisor of fractions onto limits I looked back at a task where I gave up when I got the result $\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}$ so I tried to find a way to get the $0$ out here, as well. Am I just not seeing the solution or is there no way to do it?

Hint
$$\left(\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}\cdot \frac{\sqrt{1+2x}+3}{\sqrt{x}+2}\right)\cdot \frac{\sqrt{x}+2}{\sqrt{1+2x}+3}=\frac{2x-8}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{1+2x}+3}$$
