Question

lim x+2/x^3+8
x->-2

Answer 1

Factor the denominator. For a sum of cubes, you get
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 2

L'hopstials rule?

Answer 3

@raffle_snaffle, you certainly could, but there's no guarantee it's been learned yet.

Answer 4

\[lim_{x\to -2}~\frac{x+2}{x^3+8}\]so, now we should factor the denominator, like \[x^3+y^3=(x+y)(x^2-xy+y^2)\]in this case we have\[x^3+8=(x+2)(x^2-2x+4)\]then\[lim_{x\to -2}~\frac{x+2}{(x+2)(x^2-2x+4)}\]now we can "cut" (x+2)\[\boxed{\boxed{lim_{x\to -2}~\frac{1}{(x^2-2x+4)}=\frac{1}{12}}}\]