Finding the limit of $ (1-\ln^{n+1}\lvert x\rvert)/(1-\ln\lvert x\rvert)$, as $n$ tends to infinity

Question

The problem is like this: it asks to find the set of values of the function $$ f(x) = \lim_{n\to\infty} \frac{1-\ln^{n+1}\lvert x\rvert}{1-\ln\lvert x\rvert} \qquad (x\ne0) $$ Now, what I've tried is to give some values to $x$ and it seems like for $x=1$ the limit equals $1$, for $x>1$ the limit equals infinity. I've also tried doing a graph for it and it seems like the set of values for the function is $[1,\infty)$. I was looking to see a exact method (step-by-step). I've also attached a picture.

Answer 1

Actually, the problem should also state $\ln\lvert x\rvert\ne1$, that is, $x\ne\pm e$.

You can leave out the logarithm and consider instead $$ \lim_{n\to\infty}\frac{1-t^n}{1-t}=\lim_{n\to\infty}\frac{t^n-1}{t-1} $$ where $t=\ln\lvert x\rvert$ and $t\ne1$.

Think to what you know about $$ \lim_{n\to\infty}t^n $$

1. For $t>1$, the limit is…
2. For $-1<t<1$, the limit is…
3. For $t\le -1$, the limit…

After you have filled in the dots, you will know the answer in terms of $t$ and you can then pass to $x$, because $|x|=e^t$.

Answer 2

We obtain the answer immediately:

For $|x|>e$ it goes to $+\infty$.

For $\frac{1}{e}<|x|<e$ it's closed to $$\frac{1}{1-\ln|x|}.$$ For $0<|x|\leq\frac{1}{e}$ the limit does not exist.