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: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}\times \frac{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}} = \lim_{n \to \infty}\frac{(n^2+3n+1)-(n^2+3n-1)}{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}=\lim_{n \to \infty}\frac{2}{{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}}$$

I get the limit to be $-1$.
Note that $\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)} =
\frac{ n^2( \sqrt{1+{3 \over n} + {1 \over n^2}} - \sqrt{1+{3 \over n} - {1 \over n^2}})}{n\ln(\frac{1+{1 \over n}}{1 + {2 \over n}})}$.
We have $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = -1 $ and
$\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = 1$.
To compute the first limit, let $f_0(x) = \ln(\frac{1+x}{1 + 2x})$
and note that $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = f_0'(0) = -1$.
To compute the latter limit, let $f_1(x) = \sqrt{1+{3 x} + x^2}-\sqrt{1+{3 x} - x^2}$, we have $f_1'(0) = 0, f_1''(0) = 2$, hence
$\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = {f_1''(0) \over 2} = 1$.
Hence the limit is $-1$.
