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I need to find the $\lim_{n\to\infty}\{x_n\}$ where $\{x_n\}$ is defined as $$\{x_n\}_{n\ge1}=n^{\frac{1}{n}}\;,\;n\in \mathbb{N}$$

Now if I had a function $f:\mathbb{R}-\{0\}\to\mathbb{R},\quad f(x)=x^{\frac{1}{x}}$, then I could easily find it's limit as follows: $$f(x)=x^{\frac{1}{x}}\\taking\,natural\,logarithm\,on\,both\,sides\\\ln{f(x)}=\frac{1}{x}\ln(x)\\Let\,\lim_{x\to\infty}\ln{f(x)}=L\;then\\L=\lim_{x\to\infty}\frac{1}{x}\ln(x)\\Applying\;L'Hospital's\;Rule\\L=\lim_{x\to\infty}\frac{1}{x}=0\\\Rightarrow\lim_{x\to\infty}{f(x)}=1$$

Since my sequence is clearly a subset of this function, it's limit is also 1. However, I am worried that this method may not be 'rigrous' enough. Is there any other way to find the limit of this sequence?

P.S: I tried finding the limit using the $\epsilon-N$ definition of a limit, but could not do so.

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5 Answers 5

Your method is plenty rigorous. If $f(x)\to 1$ as $x\to\infty$, then of course the sequence $f(n)$ also tends to $1$ as $n\to\infty$. Indeed, for any $\epsilon$ there exists an $M$ such that if $x>M$ then $f(x)$ is within $\epsilon$ of $1$. Thus for any $\epsilon$, pick an integer $N>M$, and we will have $f(n)$ within $\epsilon$ of $1$ whenever $n>N$, which is the definition of the limit of a sequence.

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We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see this question).

We have that $$ \lim_{n\to\infty}\frac{n+1}n=1 $$ and $$ \lim_{n\to\infty}n^{1/n}=1. $$

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Check out these answers on this site:

Find $\lim_{n\to\infty} \sqrt[n]{n}$

$\lim_{n\to\infty} \sqrt[n]n = 1$

There was another beautiful solution on this very site that used the AM-GM inequality but I can's seem to find it now. The gist of it was to write,

$$ (n)^{\frac 1 n} = (\sqrt n \cdot \sqrt n \cdot 1 \cdot 1 ... \cdot 1)^{\frac 1 n } \le \dfrac{\sqrt n + \sqrt n + 1 + 1 + . + 1 }{n} = \dfrac{2\sqrt n + n - 2 }{n} = \frac{2}{\sqrt n} - \frac{2}{n} + 1 $$

Now since $ \lim \dfrac{2}{\sqrt n} = \lim \dfrac{2}{n} = 0$, the limit on the right is $0$. Your result follows from applying the squeeze theorem.

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Alternatively, for $n > 1$ we have $n^{1/n} > 1 $. Using Bernoulli's inequality,

$$n^{1/n} = 1+a_n,\,\, (a_n > 0)\\n = (1+a_n)^n > \frac{n(n-1)a_n^2}{2}$$

Then

$$a_n < \sqrt{\frac{2}{n-1}}< \epsilon$$

for all $n$ sufficiently large, and

$$|n^{1/n}-1| = a_n < \epsilon.$$

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For your information, the method you used is completely rigorous assuming you know the following fact:

If $\lim_{x\to\infty} f(x) = L$ then the limit of the sequence $f(1)$, $f(2)$, $\dots$ is $L$.

The first limit is taken over real numbers and the second over integers. Try proving this fact if you have not seen it before. Is the converse true?

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