How do I solve this: I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases.
$$
  \lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$ 

You can multiply both numerator and denominator by $\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}$. Then you will find $$\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1}=\frac{4n^{2}+3n+2-4n^{2}-n+1}{\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}} = \frac{2n(1+3/2n)}{2n\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} =\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} $$
Thus, we have  $$\lim_{n\to \infty}\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1} = \lim_{n\to\infty}\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} = \frac{1}{2}$$
