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Let $p$ be a positive real number. How can one prove that $\displaystyle\lim_{n \to \infty} p^{\frac{1}{n}} = 1$?

I searched around the site and read this proof (using the intermediate value theorem) under the old (2014) post:

$\lim_{n \to \infty}x^{1/n}=1$ for every $x \in \mathbb{R}$ and $x>0$?

But I don't understand it:

  • I don't think that it's complete because the argument from $p=(1+q_n)^n \geq 1+nq_n \geq nq_n > 0$ only works for $p \geq 1$.

  • Why can we just say if $p^{\frac1n} = 1+q_n$? What if not?

Below is the orginal proof where I have adapted to my notation:

$f(p)=p^n$ is continuous and strictly increasing in $[0,1+p_0]$, and $$ f(0)=0<p_0<f(1+p_0)=(1+p_0)^n $$ and hence there exists a unique $q_0\in (0,1+q_0)$, such that $$ f(q_0)=q_0^n=p_0. $$

Now, if $p^{1/n}=1+q_n$, then $$ p=(1+q_n)^n\ge 1+n q_n\ge n q_n>0, > $$ and thus $$ 0<q_n\le\frac{p}{n}\to 0, $$ and hence $$ p^{1/n}=1+q_n\to 1. $$

stepbysteptomathpro
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  • Where did you see the proof? Can you cite the reference? –  Nov 27 '19 at 15:16
  • Do you mean $f(x)=p^x$ instead of $p^n$? –  Nov 27 '19 at 15:16
  • @Jack I got it from here and adapted it with $p$: https://math.stackexchange.com/questions/703117/lim-n-to-inftyx1-n-1-for-every-x-in-mathbbr-and-x0 – stepbysteptomathpro Nov 27 '19 at 15:35
  • Thanks for that. There are several answers for that post. Which one do you refer to? This one? –  Nov 27 '19 at 15:37
  • @Jack Yes, that's the one. – stepbysteptomathpro Nov 27 '19 at 15:41
  • @robjohn left a comment there that one needs to handle the case when $0<x<1$. Is your question about that? –  Nov 27 '19 at 15:46
  • @Jack Yes. I didn't want to comment there because the thread is 5 years old and no one would reply to that one – stepbysteptomathpro Nov 27 '19 at 15:47
  • OK. I have edited your post accordingly. –  Nov 27 '19 at 15:55
  • @Jack Thank you, I appreciate it. – stepbysteptomathpro Nov 27 '19 at 15:57
  • I personally don't like that proof. Using the binomial theorem is much simpler; it does not require mentioning any continuous functions. –  Nov 27 '19 at 16:00
  • And you are right, the proof there is poorly written. –  Nov 27 '19 at 16:11
  • One way to make sense of this is to first do the case where $p \geq 1$. In this case, $f(1)=1\leq p \leq (1+p)^n$ (you can even use $p\leq p^n$ for the upper bound, instead) and, via the IVT, conclude the existence of an $n$th root of $p$ between $1$ and $1+p$. In particular, this root is $\geq 1$, so it can be expressed as $1+q_n$ for some nonnegative $q_n$. Then you can conclude the argument as presented in the question. For the case where $0<p<1$, note that since $\frac{1}{p}>1$, the limit of $\frac{1}{p}^{\frac{1}{n}}$ as $n$ goes to infinity will be $1$. Conclude from there. – CardioidAss22 Nov 27 '19 at 16:28

2 Answers2

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In comments, you mentioned that your question was regarding the fact that the proof cited only works for $p\ge1$. If $p\lt 1$, then $\frac1p\gt1$ and the proof there shows that $$ \lim_{n\to\infty}\left(\frac1p\right)^{1/n}=1 $$ Since $\frac1x$ is continuous at $x=1$, we can then say that $$ \lim_{n\to\infty}p^{1/n}=\frac11=1 $$

robjohn
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Using the continuity of the exponential function : $\lim_{n\to\infty} p^{\frac{1}{n}} = \lim_{n\to\infty} \exp(\frac{1}{n}\ln p)= \exp(\lim_{n\to\infty}\frac{1}{n}\ln p) = \exp(0)=1$.

Mister Da
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