limit as n goes to infinity of $\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$ 
How do you go about solving $$\lim_{n\to\infty}\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$$

I know that I have to fix the top so that it is not $(\infty - \infty$),
but if I multiple it by 
$$\frac{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}}{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}},$$
the bottom part becomes very ugly and extremely hard to deal with.
 A: This is a straightforward calculation: Expanding the fraction by $\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}$ gives
\begin{align*}
 &\, \frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}} \\
=&\, \frac{(\sqrt{n^3-3}-\sqrt{n^3+2n^2+3)}(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})}{\sqrt{n+2} (\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})} \\
=&\, \frac{(n^3-3)-(n^3+2n^2+3)}{\sqrt{(n+2)(n^3-3)} + \sqrt{(n+2)(n^3+2n^2+3)}} \\
=&\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}}.
\end{align*}
Dividing numerator and denumerator by $n^2$ gves
\begin{align*}
  &\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}} \\
 =&\, \frac{-2-\frac{6}{n^2}}{\sqrt{1+\frac{2}{n}-\frac{3}{n^3}-\frac{6}{n^4}} + \sqrt{1+\frac{4}{n}+\frac{4}{n^2}+\frac{3}{n^3}+\frac{6}{n^4}}}.
\end{align*}
Taking the limit $n \to \infty$ results in $-2/2 = -1$.
A: When you are at the stage
$$
\lim_{n\to\infty}\frac{-2n^2-6}{
  \sqrt{n+2}\,(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})
}
$$
pull $n^2$ from the numerator, $n$ from $n+2$ and $n^3$ from the other two radicals, so you have
$$
\lim_{n\to\infty}\frac{n^2\left(-2-\dfrac{6}{n^2}\right)}{
  \sqrt{n}\,\sqrt{1+\dfrac{2}{n}}\,
  \sqrt{n^3}\left(
    \sqrt{1-\dfrac{3}{n^3}}+\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^3}}
  \right)
}
$$
Cancel $n^2$ with $\sqrt{n}\,\sqrt{n^3}$ and you're done.
