<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] For the sake of being concise, I have chosen to omit most of the intermediary algebraic simplifications. Letting $u = x+2$, we get 
$$\lim_{u \to 4} \frac{{u^{\frac 52} - 4u^{\frac 32} + 4\sqrt{u} - 8}}{4 - u^2 + 4u - 4}$$
By L'hopital, this is equal to 
$$\lim_{u \to 4} \frac{\frac {5u^2 - 12u+4}{2 \sqrt{u}}}{4 - 2u} = \lim_{u \to 4} \frac{2 - 5u}{4\sqrt{u}} = -\frac{9}{4}$$
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