<s>[INST] limit of an indetermined sequence I'm trying to find out the limits of a sequence of the type $\frac{\infty}{\infty}$, but I got stuck and am starting to get frustrated. The sequence in question is: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}$$ I tried looking at its parts and figuring out their own limits to try and help me, but I didn't get anywhere useful. I tried rationalizing the square roots and this is what I have at the moment: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}\times \frac{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}} = \lim_{n \to \infty}\frac{(n^2+3n+1)-(n^2+3n-1)}{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}=\lim_{n \to \infty}\frac{2}{{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}}$$
 [/INST] Observe that
$$
\lim_{n\to\infty}(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})=
\lim_{n\to\infty}\frac{2}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=0
$$
as well as
$$
\lim_{n\to\infty}(\ln(1+n)-\ln(n+2))=
\lim_{n\to\infty}\ln\frac{1+n}{n+2}=0
$$
so your limit is in the form $0/0$, rather than $\infty/\infty$.
On the other hand, you can observe that
$$
\lim_{n\to\infty}n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})=
\lim_{n\to\infty}\frac{2n}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=1
$$
and that
$$
\lim_{n\to\infty}n(\ln(1+n)-\ln(n+2))=
\lim_{n\to\infty}n\ln\frac{1+n}{n+2}
$$
is finite as well: with $t=1/n$, the limit becomes
$$
\lim_{t\to0^+}\frac{\ln(t+1)-\ln(1+2t)}{t}=-1
$$
So you can write your limit as
$$
\lim_{n \to \infty}\frac{n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})}{n(\ln(1+n) - \ln(2+n))}
$$
and finish up.
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