Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$

Substitute $t=\sqrt{x+2}$; if $x\to2$, then $t\to2$, so, after doing $x=t^2-2$, you get
\begin{align}
\lim_{t\to2}\frac{t(t^2-2)^2-8}{4-(t^2-2)^2}
&=\lim_{t\to2}\frac{t^5-4t^3+4t-8}{4t^2-t^4} \\[4px]
&=\lim_{t\to2}\frac{t^3(t^2-4)+4(t-2)}{t^2(4-t^2)} \\[4px]
&=\lim_{t\to2}\left(-t-\frac{4}{t^2(2+t)}\right)
\end{align}
