Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$ 
Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}

I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.
I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work. 
I have tried to multiply by the conjugates of the numerator only but it did not work.
So what shall I do?

Rationalise the numerator.
$\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\right)=\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\times\dfrac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3}\right)=\lim_{x\rightarrow4}\left(\dfrac{2(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}\right)=\dfrac{2(\sqrt{4}+2)}{\sqrt{1+2(4)}+3}=\dfrac{8}{9}=\dfrac43$
