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.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$

Since the OP does not know L'Hospital's rule (yet), this approach is probably not the best way. When you multiply top and bottom by $x^2\sqrt{x+2}+8$ you get the fraction: $\frac{x^5+2x^4-64}{(4-x^2)(x^2\sqrt{x+2}+8)}$ Both NUM and DENOM have a factor $x-2$ In case of the NUM, you could perform long division to arrive at $x^4+4x^3+8x^2+16x+32$ and in case of the DENOM, it is of course straight forward. Can you finish it from here?
