prove: $\lim_{n \rightarrow \infty} x^{1/n} = 1$

Question

I have to prove that $\lim_{n \rightarrow \infty}$ $x^{1/n} = 1$ for $x > 0$.

I splitted it up in 3 cases:

$x = 1:$
$1^{1/n} = 1$ $\forall$ $n$, so $\lim_{n \rightarrow \infty}$ $x^{1/n} = 1$ if $x =1$

$x > 1:$
I already proved that $(x^{1/n})_{n = 1}^{\infty}$ is decreasing if $x > 1$, so I only have to prove that it's infimum is 1. How do I prove that?

$x < 1:$
I don't know how the proof works in this case.

Could you please help me completing the second and third case?

Thanks in advance!

For the last case, set $x=\dfrac 1y,\enspace y>1$. What is infinum? — Bernard, Mar 15 '15 at 13:27
This must have been asked countless times before at this site... Here's one special case at least: http://math.stackexchange.com/questions/125588/limit-of-21-n-as-n-to-infty-is-1/ — Hans Lundmark, Mar 15 '15 at 13:30
@Hans Lundmark No, it's not the same. That's just a special case of case 2 — Peter, Mar 15 '15 at 13:31
What should I use? Bernoulli's inequality or Binomial theorem or L'hospital or Logarithm or just crude proof, what's your order? — RE60K, Mar 15 '15 at 13:32
@ADG I think I will understand the crude proof the best if you don't mind — Peter, Mar 15 '15 at 13:33
Yes, so it's a special case! (Of case 2, not case 3, but nevermind...) Anyway, Bernard's comment already tells you how to take care of case 3 once you know how to do case 2. — Hans Lundmark, Mar 15 '15 at 13:34
As suggested already many times, the high road is to consider the logarithm, and to use the continuity of the exponential function. A somewhat more elementary approach shows first that the sequence is monotonous then solves graphically the limit of $x^{1/2^n}$, drawing the functions $t\mapsto t$ and $t\mapsto\sqrt{t}$ on the same diagram and drawing the usual stairs to iterate the latter. — Did, Mar 15 '15 at 13:38
@HansLundmark Have you downvoted the question? If it was you, it's very childisch. I did not know that someone proved a special case of case 2 before. I didn't see it on google — Peter, Mar 15 '15 at 13:38
No, I didn't downvote, and I didn't even vote to close as a duplicate. I just pointed you to another question with several good answers which actually should solve half of your problem (just change "2" to "x"). (And as I said, Bernard's comment then solves the other half.) — Hans Lundmark, Mar 15 '15 at 15:20
@HansLundmark OK sorry. Then it was Bernard. Thanks for your help — Peter, Mar 15 '15 at 15:37
Possible duplicate of Prove $\forall K > 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$ — , Nov 27 '19 at 16:21
Possible duplicate of Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$ — Arnaud D., Dec 03 '19 at 10:17

RE60K · Answer 1 · 2015-03-15T13:58:44.520

2

I don't think whether this would be a standard proof,

but when $n\to\infty,1/n\to0$ so $x^{1/n}\to x^{0}=1\forall x>0$

edited Mar 15 '15 at 13:58

answered Mar 15 '15 at 13:38

RE60K

17,716

Thank you! I appreciate it very much that you are trying to help. But maybe Bernoulli's inequality would be better. Could you please do that for me? – Peter Mar 15 '15 at 13:42
@Peter see the edit. – RE60K Mar 15 '15 at 13:46
Thank you! that's great. I understand it now – Peter Mar 15 '15 at 13:47
@ADG I don't know if I'm not getting it or so, but how does the Bernoulli one help compute the limit? It seems to me to only mean that the limit is greater than 1. (Considering that $r\to0$) – Hasan Saad Mar 15 '15 at 13:48
@HasanSaad when does equality hold in this inequality? – RE60K Mar 15 '15 at 13:49
I don't know, but it seems to me something might be wrong, but maybe it's just me. That inequality, as you wrote it, holds only for $r\geq1$ or $r\leq0$. (1+2)^0.5=1.73205081 1+2(0.5)=2 – Hasan Saad Mar 15 '15 at 13:54
Well you need continuity of the function $t\mapsto x^t$ for your argument to work. – Surb Mar 15 '15 at 14:08

score 2 · Answer 2 · answered Mar 15 '15 at 13:58

2

For $x>1$ Bernouilli's inequality yields $$\left(1+\frac{(x-1)}{n}\right)^n\ge1+n\frac{(x-1)}{n}=x,$$ so $$ \left(1+\frac{(x-1)}{n}\right)\ge x^{\frac1n}\ge1.$$ Now let $n$ tend to infinity.

answered Mar 15 '15 at 13:58

Vincent Boelens

3,814

@kobe This is true, but $x^{\frac1n}\ge1$ is only true if $x\ge 1$ – Vincent Boelens Mar 15 '15 at 14:19

Hasan Saad · Accepted Answer · 2015-03-15T16:12:56.307

1

The case where $a=1$ is trivial.

We know that $\lim_{n\to\infty}\frac{a}{q^n}=0$ when $q>1$

Let $a>1$ and let $q=1+\varepsilon$

Thus, for some $N$, $1<a<(1+\varepsilon)^n$ for all $n>N$

Then $1<a^{\frac{1}{n}}<1+\varepsilon$ for all $n>N$

Thus, by the definition of the limit $\lim_{n\to\infty}a^\frac{1}{n}=1$

Now, if $a<1$, then letting $b=\frac{1}{a}$

$\lim_{n\to+\infty} a^\frac{1}{n}=\frac{1}{\lim_{n\to\infty}b^\frac{1}{n}}=1$

QED.

As for the proof of $\lim_{n\to\infty}\frac{a}{q^n}=0$, I shall write it here.

Let $n>log_q(\frac{a}{\varepsilon})$

Then $q^n>\frac{a}{\varepsilon}$

And thus, $\frac{a}{q^n}<\varepsilon$

Also, if you can't use logarithms, this proof is evident from the principle(or axiom) of Archimedes

edited Mar 15 '15 at 16:12

answered Mar 15 '15 at 13:41

Hasan Saad

4,545

Actually, the proof for $a>1$ includes $a=1$ actually but I just made them into $3$ parts because you did. – Hasan Saad Mar 15 '15 at 13:42
Thank you! your answer is very helpful, same as ADG's – Peter Mar 15 '15 at 13:44
I was only wondering if I am allowed to use that $\lim_{n\to\infty}\frac{a}{q^n}=0$, or is there a simple proof for that? – Peter Mar 15 '15 at 15:55
I think you should be allowed to use it as it is honestly the simplest thing... As for proving it, I'll edit my answer to contain the proof. – Hasan Saad Mar 15 '15 at 16:04

score 1 · Answer 4 · answered Nov 27 '19 at 16:21

1

$$\log(x^{1/n})=\frac{\log(x)}n\to0$$

and as the exponential is a continuous function,

$$\lim_{n\to\infty}e^{\log(x)/n}=e^{\lim_{n\to\infty}\log(x)/n}.$$

answered Nov 27 '19 at 16:21

prove: $\lim_{n \rightarrow \infty} x^{1/n} = 1$

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