Limit $\lim_{n\to\infty} (1+x^{n})^{\frac{1}{n}}$

Question

Question: I can't seem to show that the limit of $(1+x^{n})^{\frac{1}{n}}$ as $n\rightarrow\infty$ is $x$, where $x$ is in the interval $[1,2]$.

Attempt: Let $y_{n}=(1+x^{n})^{\frac{1}{n}}$.

Then $\lim_{n\rightarrow\infty}{y_{n}}=\lim_{n\rightarrow\infty}{(1+x^{n})^{\frac{1}{n}}}$.

Taking logarithms we have $\lim_{n\rightarrow\infty}{\ln{y_{n}}}=\lim_{n\rightarrow\infty}{\frac{1}{n}\ln{(1+x^{n})}}$.

Here I decided to substituted $v=1/n$, so the limit is now taken as $v$ goes to $0$.

Then $\lim_{n\rightarrow\infty}{\ln{y_{n}}}=\lim_{v\rightarrow{0}}{v\ln{(1+x^{\frac{1}{v}})}}$.

Now the RHS is equivalent to $\lim_{v\rightarrow{0}}{v}\cdot\lim_{v\rightarrow{0}}{\ln{(1+x^{\frac{1}{v}})}}$.

The problem now is that this becomes the indeterminate $0\cdot\infty$.

Any help will be appreciated :)

Hint: $1 + x^n = x^n\left(1 + \frac{1}{x^n}\right)$. – Daniel Fischer Oct 12 '13 at 17:00 — Daniel Fischer, Oct 12 '13 at 17:00
@DanielFischer thanks, i managed to show it now. – WhizKid Oct 12 '13 at 17:14 — WhizKid, Oct 12 '13 at 17:14

score 4 · Accepted Answer · edited May 04 '24 at 06:34

4

Since $1\leq x\leq 2$, then $x^n\leq 1+x^n\leq 2x^n$, that is, $x\leq (1+x^n)^{\frac{1}{n}}\leq 2^{\frac{1}{n}}x$. therefore, $\lim_{n\to\infty}(1+x^n)^{\frac{1}{n}}=x$.

edited May 04 '24 at 06:34

jimjim

9,855

answered Oct 12 '13 at 17:24

Lei Li

626

Nice proof. All you seem to need is $x \geq 1$: $x \leq 2$ is never used. – Stefan Smith Oct 12 '13 at 17:51
Yes. All my need is $x\geq 1$. So the limit is true for all $x\geq 1$. – Lei Li Oct 13 '13 at 08:50

score 2 · Answer 2 · answered Oct 12 '13 at 19:10

Another approach with logarithms: put

$$f(y):=\left(1+x^y\right)^{1/y}\implies \log f(y)=\frac{\log(1+x^y)}{y}$$

and applying l'Hospital:

$$\lim_{y\to\infty}\log f(y)=\lim_{y\to\infty}\frac{\log(1+x^y)}y=\lim_{y\to\infty}\frac{x^y}{1+x^y}\log x=\log x\implies$$

$$\lim_{y\to\infty}f(y)=\lim_{y\to\infty}e^{\log f(y)}=e^{\log x}=x$$

and since the above's true for the continuous variable $\;y\;$ then it is true for any way we choose to make it go to infinity, and in particular for $\;y=n\in\Bbb N\;$

user · Answer 3 · 2024-05-04T07:33:49.540

0

As an alternative, by ratio-root criterion with $y_{n}=a_n^{\frac{1}{n}}=(1+x^{n})^{\frac{1}{n}}$

$$\frac{a_{n+1}}{a_n}=\frac{1+x^{n+1}}{1+x^{n}}=x\frac{\frac 1{x^{n+1}}+1}{\frac 1{x^{n}}+1}\to x$$

edited May 04 '24 at 07:33

answered May 04 '24 at 07:01

user

162,563

Limit $\lim_{n\to\infty} (1+x^{n})^{\frac{1}{n}}$

3 Answers3