L'Hospital's Rule Indeterminate Form Question I'm trying to solve this limit, but I don't know if I can use L'Hopital's Rule on it in the current form.
$$\lim_{n \to \infty} {\ln ({3n+1}) - \ln ({3n-1}) \over \frac 1n} $$
I tried to look what $\infty - \infty$ is equal to, but it seems that it varies based on what type of infinity these are, so that didn't help. If this doesn't work I don't see any other way to get it into a form that works, but if I am wrong pointers in the right direction would be useful.
Please don't solve the problem for me, this the middle of a homework problem so I am trying to figure this out at least partially on my own. Thanks
 A: HINT :
Use the subtraction property of logarithm for the expression in the numberator and you'll get :
$$\lim_{x \to \infty} {\ln ({3n+1}) - \ln ({3n-1}) \over \frac 1n}= \lim_{x \to \infty} \frac{\ln\bigg(\frac{3n+1}{3n-1}\bigg)}{\frac{1}{n}}$$
Now this is an intermediate form of $\frac{0}{0}$ when $x \to \infty$ so you can proceed with applying L'Hospital's Rule. 
A: Just another way
As Rebellos answered
$$A=\lim_{x \to \infty} {\ln ({3n+1}) - \ln ({3n-1}) \over \frac 1n}= \lim_{x \to \infty} \frac{\ln\bigg(\frac{3n+1}{3n-1}\bigg)}{\frac{1}{n}}=\lim_{x \to \infty} \frac{\ln\bigg(1+\frac{2}{3n-1}\bigg)}{\frac{1}{n}}$$ Now, using equivalents
$$\ln\bigg(1+\frac{2}{3n-1}\bigg)\sim \frac{2}{3n-1}$$ making 
$$A\sim \frac{2n}{3n-1}\to \,\,?$$
A: You can also not use L'Hospital rule doing as follows
\begin{eqnarray}
\lim_{n\to\infty}\frac{\ln(3n+1)-\ln(3n-1)}{1/n}&=&
\lim_{n\to\infty}\ln\left(\frac{3n+1}{3n-1}\right)^{n}\\&=&\lim_{n\to\infty}\ln\left(\frac{1+\frac{1}{3n}}{1-\frac{1}{3n}}\right)^{n}\\&=&\lim_{n\to\infty}\ln\frac{\left(1+\frac{1}{3n}\right)^{n}}{\left(1-\frac{1}{3n}\right)^{n}}.
\end{eqnarray}
Now use that $\lim_{n\to \infty}\left(1+\dfrac{\alpha}{n}\right)^n=e^\alpha$.
A: By Taylor's series for natural logarithm
$$\log(1+x)=x+o(x)$$
we have
$${\ln ({3n+1}) - \ln ({3n-1}) \over \frac 1n} = n\left(\ln \left(1+\frac1{3n}\right) - \ln \left(1-\frac1{3n}\right)\right)=n\left(\frac1{3n}+\frac1{3n}+o\left(\frac1n\right)\right)=\frac23+o(1)\to\frac23$$
