How can one prove that $\lim_{n \to \infty}x^{1/n}=1$ for every $x>0$?

Question

Two questions:

1) Let $x \in \mathbb{R}$ and $y \in \mathbb{R}$ and $x>0$ and $y>0$. Let $n \in \mathbb{N}$. We know that

For every real $x>0$ and every integer $n>0$ there is one and only one real $y>0$ such that $y^n=x$. The number $y$ is written $x^{1/n}$. (Rudin, 1976, p.10 with addition of $y>0$)

Is it correct to affirm that

$$\lim_{n \to \infty}x^{1/n}=1$$

for any $x$?

2) When I write $n \to \infty$ I can affirm that $n<\infty$?

$x^{1/n}$ is undefined for $x\leq 0$ unless you are talking about complex exponential functions. — , Mar 07 '14 at 15:49
When I affirm that $n \to \infty$ with the meaning that $n$ goes to $\infty$ it means that $n$ goes towards $\infty$ but it will never reach $\infty$ and so $n < \infty$ @robjohn — Erivelton Geraldo Nepomuceno, Mar 07 '14 at 15:55
Possible duplicate of Prove $\forall K > 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$ — , Nov 27 '19 at 16:20
Possible duplicate of Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$ — Arnaud D., Dec 03 '19 at 10:17

score 2 · Answer 1 · edited Nov 27 '19 at 16:09

2

The simplest way to show this fact is using the Intermediate Value Theorem: If $f:[a,b]\to \mathbb R$ is continuous, and $f(a)<\xi<f(b)$, then there exists a $c\in(a,b)$, such that $f(c)=\xi$. Further, if $f$ is strictly increasing, then the $c$ is unique.

So, here $f(x)=x^n$ is continuous and strictly increasing in $[0,1+x_0]$, and $$ f(0)=0<x_0<f(1+x_0)=(1+x_0)^n $$ and hence there exists a unique $y_0\in (0,1+x_0)$, such that $$ f(y_0)=y_0^n=x_0. $$

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

While $n\to\infty$, the value of $n$ is positive, eventually larger than any positive numbers, but still a number and not infinite.

edited Nov 27 '19 at 16:09

answered Mar 07 '14 at 15:36

Yiorgos S. Smyrlis

83,933

And regarding the second question, do you have a position? – Erivelton Geraldo Nepomuceno Mar 07 '14 at 15:40
Thanks for answering the second question. – Erivelton Geraldo Nepomuceno Mar 07 '14 at 15:58
Your answer seems perfect. But does it contradict the other answer below? @YiorgosS.Smyrlis – Erivelton Geraldo Nepomuceno Mar 07 '14 at 16:03
1

The argument from $x=(1+y_n)^n\ge 1+n y_n\ge n y_n>0$ only works for $x\ge1$. It is easy to extend this result, but it should be handled. – robjohn Mar 07 '14 at 20:36
@robjohn: a question about this answer has been just posted by a user. – Nov 27 '19 at 15:48

robjohn · Answer 2 · 2014-03-07T17:58:19.213

1) Bernoulli's Inequality says that for all $n\ge1$ and $u\ge-1$, $$ 1+nu\le(1+u)^n\tag{1} $$ Let $x=1+nu$ and take $n^\text{th}$ roots: $$ x^{1/n}\le1+\frac{x-1}{n}\tag{2} $$ Taking reciprocals of $(1)$: $$ \frac1{1+nu}\ge\left(1-\frac{u}{1+u}\right)^{\large n}\tag{3} $$ Let $x=\frac1{1+nu}$ and take $n^\text{th}$ roots: $$ x^{1/n}\ge1-\frac{1/x-1}{1/x-1+n}\tag{4} $$ Using $(2)$ and $(4)$ and the Squeeze Theorem gives $$ \lim_{n\to\infty}x^{1/n}=1\tag{5} $$ 2) since $\infty$ is not a real number, $n$ will never be $\infty$. When we write $\lim\limits_{n\to\infty}\,$, we simply mean that $n$ gets arbitrarily big, not that it ever becomes infinite.

Jose Antonio · Answer 3 · 2014-03-08T03:03:02.970

One good way to obtain the result is divide the proof. First try to show that for every $M>0$ and every $\varepsilon>0$ there is a $n$ such that $M^{1/n}\le 1 + \varepsilon$ (Hint: use the fact that if $x>1$, then $x^n$ diverges). After this you may treat the cases $x\ge 1$ and $x<1$ separately.

With respect to the last point of the question number 1) you can also define the $n$-th root as follows, let $n \ge 1$, $x>0$ and $E= \{z\in \mathbb{R}^{\ge0}:z^{n}\le x\}$. Then $x^{1/n}: = \sup E$.

We need to show that this definition make sense.

Existence: First of all we have to show that least upper bound property can be apply. Clearly the set is non-empty because $0 \in E$. So only we need to show that the set is bounded above. We divide the proof in two cases, $x\le1$ and $x>1$. For the first case, we claim that $1$ is an upper bound. If were not the case, there is an element $y\in E$ such that $y>1$, then $y^n>1$, $y^n > x$ and hence $y^n \notin E$, a contradiction. Now suppose that we are in the case $x>1$, we claim that $x$ itself is an upper bound. If were not the case, there is some element $y\in E$ such that $y>x$ so $y^n>x^n$, and since $x>1$ we have $x^n\ge x$. So $y^n>x$, $y \notin E$, a contradiction. Thus in both cases $E$ has an upper bound.

Now we have to show that this definition has the desired property. In other words, if $y =x^{1/n}$, then $y^n=x$

Suppose to the contrary that $y^n<x$. Choose $\varepsilon<\min \bigg(y, \frac{x-y^n}{2^ny^{n-1}}\bigg)$. Then we have

$$(y+\varepsilon)^n<y^n+2^n\varepsilon y^{n-1}<x$$

Then $y+\varepsilon \in E$ which contradicts that $y$ is an upper bound. A very similar argument can be used to show that $y^n>x$ also yields a contradiction (in this case, choosing $\varepsilon>0$ small enough such that $(y-\varepsilon)^n>x$, contradicting that $y$ is the least upper bound). Thus the only alternative is $y^n=x$ as desired.

Notice that the uniqueness follows immediately from the definition of least upper bound.

score 0 · Answer 4 · answered Mar 08 '14 at 02:51

The fact that $\lim_{n\to \infty} x^{1/n}=1$ can be shown in two easy steps:

Show that the sequence $\{x^{1/n}\}$ is either strictly increasing and bounded above by $1$, if $0<x\leq 1$, or strictly decreasing and bounded below by $1$, if $x>1$. Then use the monotone convergence theorem to prove that the sequence has a limit.
Let the limit be $L$. Then, the limit of the sequence $\{\log(x^{1/n})\}$ must be $\log L$, because $\log $ is continuous on $(0,\infty)$. Since $$ \log(x^{1/n})=\frac{\log x}{n} \to 0,$$ as $n\to \infty$, it follows that $\log L=0$, and therefore $L=1$.

How can one prove that $\lim_{n \to \infty}x^{1/n}=1$ for every $x>0$?

4 Answers4

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