Calculate $\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}$

Question

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?

Answer 1

Hint

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

Answer 2

With $\sqrt x=t+2$, the limit becomes

$$\lim_{t\to0}\frac{\sqrt{9+8t+2t^2}-3}t.$$

Then multiplying/dividing by the conjugate,

$$\lim_{t\to0}\frac{8t+t^2}t\frac1{\sqrt{9+8t+2t^2}+3}=8\frac1{3+3}.$$