Calculate the limit not using L'Hospitale rule:
$\lim\limits _{x \to 2} \frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$
I tried multiplying for both conjugates but I got nothing. I think I have to make something with $\sqrt{x^3-4}\cdot \sqrt{x^2+12}$, but I don't know what to do.
\begin{align} \lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}&=\lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}\color{blue}{\cdot\frac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4}\cdot\frac{2+\sqrt{x^3-4}}{2+\sqrt{x^3-4}}}\\ &=\lim_{x\to2}\frac{2+\sqrt{x^3-4}}{\sqrt{x^2+12}+4}\cdot\frac{(x^2+12)-(16)}{(4)-(x^3-4)}\\ &=\frac12\cdot\lim_{x\to2}\frac{x^2-4}{8-x^3} \end{align}
Can you take it from here?