<s>[INST] 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.
$$
  \lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$ 
 [/INST] $$\left({\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}}\right)\frac{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}\\=\frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
so evaluate 
$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}+\\\lim_{n\to\infty} \frac{3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
by which
$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
where the second term is zero because the numerator is a constant while the denominator contains $\sqrt{4n^2}$. Therefore
\begin{align}\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}&=\lim_{n\to\infty} \frac{2n}{{2n\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+2n\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\lim_{n\to\infty} \frac{1}{{\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\frac{1}{2}
\end{align}
</s>