<s>[INST] 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.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
 [/INST] L'Hôpital's rule is not necessary. Here are the steps
$$\lim\limits_{x\to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
$$=\lim\limits_{x\to 2} \frac{8-x^2\sqrt{x+2}}{(x-2)(x+2)}$$
Let $t=\sqrt{x+2}$, then
$$\lim\limits_{t\to 2} \frac{8-\left(t^2-2\right)^2 t}{\left(t^2-4\right)t^2}$$
$$=\lim\limits_{t\to 2} \frac{\left(-t^5+4t^3-4t+8\right)}{(t-2)(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-(t-2)\left(t^4+2t^3+4\right)}{(t-2)(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-\left(t^4+2t^3+4\right)}{(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-\left(t^2+2t+\frac{4}{t^2}\right)}{t+2}$$
$$=-\frac{\left(4+4+1\right)}{2+2}$$
$$=-\frac{9}{4}$$
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