<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] It seems you may not know l'Hôpital's rule. It's also possible without the need for differentiation. Multiply numerator and denominator with the so called conjugate expression of the numerator:
$$\frac{x^2 \sqrt{x+2} - 8}{4-x^2}
= \frac{\left( x^2 \sqrt{x+2} - 8 \right)\left( x^2 \sqrt{x+2} + 8 \right)}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$
Then simplify via $(a-b)(a+b) =a^2-b^2$ in the numerator:
$$\frac{x^4(x+2)-64}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$
Now you should factorise so you have a common factor $x-2$ in the numerator and denominator. In the denominator that's easy via $4-x^2 = (2-x)(2+x)=-(x-2)(x+2)$. For the nominator, you know that the polynomial divides $(x-2)$ since $x=2$ is a root of the polynomial. You can use long division or Horner's method for this:
$$x^4(x+2)-64 = x^5+2x^4-64=(x-2) (x^4+4 x^3+8 x^2+16 x+32)$$
So back to the limit:
$$\lim_{x \to 2} \frac{x^2 \sqrt{x+2} - 8}{4-x^2}
=\lim_{x \to 2}  \frac{(x-2) (x^4+4 x^3+8 x^2+16 x+32)}{-(x-2)(x+2) \left( x^2 \sqrt{x+2} + 8 \right)}$$
Cancel the common factor $x-2$ and plug in $x=2$, simplify:
$$\lim_{x \to 2}  \frac{x^4+4 x^3+8 x^2+16 x+32}{-(x+2) \left( x^2 \sqrt{x+2} + 8 \right)} = \ldots = -\frac{9}{4}$$
If you have seen l'Hôpital's rule, the other answers give you an alternative.
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