<s>[INST] Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2 
$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$

I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and delighted if someone could give some hints about how to atack this kind of problem... what is the thinking behind? Is it just about intuition? Or is it about test all sort of algebraic manipulation?  
 [/INST] You need to get used to the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which solves these algebraic limit problems instantly.
Thus we have
\begin{align}
L &= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - 4}{2 - \sqrt{x^{3} - 4}}\notag\\
&= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - \sqrt{16}}{(x^{2} + 12) - 16}\cdot\frac{x^{2} - 4}{8 - x^{3}}\cdot\frac{4 - (x^{3} - 4)}{\sqrt{4} - \sqrt{x^{3} - 4}}\notag\\
&= -\lim_{t \to 16}\frac{t^{1/2} - 16^{1/2}}{t - 16}\cdot\lim_{x \to 2}\frac{x^{2} - 2^{2}}{x - 2}\cdot\frac{x - 2}{x^{3} - 2^{3}}\cdot\lim_{u \to 4}\frac{u - 4}{u^{1/2} - 4^{1/2}}\text{ (}u = x^{3} - 4, t = x^{2} + 12)\notag\\
&= -\frac{1}{2}\cdot 16^{-1/2}\cdot 2\cdot 2^{1}\cdot\frac{1}{3\cdot 2^{2}}\cdot\dfrac{1}{\dfrac{1}{2}\cdot 4^{-1/2}}\notag\\
&= -\frac{1}{3}\cdot\frac{2}{4}\notag\\
&= -\frac{1}{6}\notag
\end{align}
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