L'Hospital's Rule Indeterminate Form Question

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

Answer 1

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.

Answer 2

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 $\lim_{n\to \infty}\left(1+\dfrac{\alpha}{n}\right)^n=e^\alpha$.

Answer 3

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 $$A\sim \frac{2n}{3n-1}\to \,\,?$$

Answer 4

By Taylor's series for natural logarithm

$$\log(1+x)=x+o(x)$$

we have

$${\ln ({3n+1}) - \ln ({3n-1}) \over \frac 1n} = n\left(\ln \left(1+\frac1{3n}\right) - \ln \left(1-\frac1{3n}\right)\right)=n\left(\frac1{3n}+\frac1{3n}+o\left(\frac1n\right)\right)=\frac23+o(1)\to\frac23$$