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.
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 $-2/2 = -1$.
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.
