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?

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