Does the limit $\lim_{n\to\infty}\left(x^n-1\right)^{1/n}$ exist?

Question

For $x$ given, what do you think about the following limit?

$$ \lim_{n\to\infty}\left(x^n-1\right)^{1/n}. $$

What I tried and what are the problems that I am facing:

Let $f(x, n)=\left(x^n-1\right)^{1/n}$. We have:

$$ \log f(x, n)=\dfrac{1}{n}\log\left(x^n-1\right)=\dfrac{1}{n}\log\left(1-x^{-n}\right)+\dfrac{1}{n}\log\left(x^n\right), $$

first, I do not know if I can apply the log or not? I guess $x$ must be real? and must be positive? what about complex?

Finally, $$ \lim_{n\to\infty}\left(x^n-1\right)^{1/n}=\log x. $$

I would imagine that this problem only makes sense when $x$ is a positive real number, because the $n$th root is not well-defined if $x^n-1$ is not positive. — Lee Mosher, Jun 10 '14 at 16:10
Also, what you have proved is that $\lim_{n\to\infty} \log f(x,n) = \log(x)$, not $\lim_{n\to\infty} f(x,n) = \log(x)$. — Lee Mosher, Jun 10 '14 at 16:12

Rene Schipperus · Accepted Answer · 2014-06-10T16:30:15.960

7

Here is a down and dirty solution, for large $n$, (and assuming $x>1$, since otherwise how can you take $n$ th root),

$$\frac{1}{2}x^n \leq x^n-1 \leq x^n$$
so

$$\frac{1}{\sqrt[n]{2}}x \leq (x^n-1)^{\frac{1}{n}} \leq x$$
So $$(x^n-1)^{\frac{1}{n}} \to x$$

edited Jun 10 '14 at 16:30

answered Jun 10 '14 at 16:12

Rene Schipperus

39,526

Excellent answer! It should be noted that $\lim\limits_{n\to\infty}\sqrt[n]{a}=1$, so we are using the squeeze theorem here. – Alex Schiff Jun 10 '14 at 16:27
To editors well obviously $a$ must be positive, I thought I was pathetic spending all my time aswering these questions, but I guess you are doing me one better. – Rene Schipperus Jun 10 '14 at 16:27
I spend time answering the questions I can, there are so many technical questions on the site that are way above my head. – Alex Schiff Jun 10 '14 at 16:28
@AlexSchiff yes, I changed my answer a little, your remark made me think that you can just use a fixed $a$, in the limit you give. – Rene Schipperus Jun 10 '14 at 16:31
@AlexSchiff Well sorry if I was harsh, it just seems to me that editing other people questions and answers is a waste of time, if they write something I dont like, or is wrong, I just make a comment, I think it is polite. – Rene Schipperus Jun 10 '14 at 16:35
Sorry Rene, didn't mean to offend. I thought it was a small change so I didn't think it would be a big deal. – Alex Schiff Jun 10 '14 at 17:31
No its Ok I am not mad at you. – Rene Schipperus Jun 10 '14 at 17:34
This is a really nice solution. simple and clear. +1 – Paramanand Singh Jun 11 '14 at 03:46

score 0 · Answer 2 · answered Jun 11 '14 at 21:48

0

Another approach if I may, just for fun:

$\lim_{n\to\infty}\left(x^n-1\right)^{1/n}=e^{ln(\lim_{n\to\infty}\left(x^n-1\right)^{1/n})}=e^{\lim_{n\to\infty}\frac{ln\left(x^n-1\right)^{1/n}}{n}}=e^{\lim_{n\to\infty}\frac{\left(x^nln(x)\right)}{x^n-1}}=e^{\lim_{n\to\infty}\frac{\left(x^n(ln(x))^2\right)}{x^n(lnx)}}=e^{\lim_{n\to\infty}lnx}=e^{lnx}=x$

answered Jun 11 '14 at 21:48

user114138

897

That is the approach OP gave, in disguise – vonbrand Jun 12 '14 at 00:37
I know... still won't hurt and might help others – user114138 Jun 12 '14 at 08:35

Does the limit $\lim_{n\to\infty}\left(x^n-1\right)^{1/n}$ exist?

2 Answers2