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When I evaluate the limit in the title above I get the following:

\begin{align} \lim\limits_{n\to\infty}\dfrac{1}{\sqrt[n]{n}} &= \lim\limits_{n\to\infty} \dfrac{1}{n^{\frac{1}{n}}} = \dfrac{1}{\infty^0} \quad\Rightarrow\quad Indeterminate\\ &= \lim\limits_{n\to\infty}\left(\dfrac{1}{n}\right)^\frac{1}{n} = 0^0 \quad\Rightarrow\quad Indeterminate \end{align}

But when I use a computer software (mathematica) to evaluate the same limit it says the limit is 1. What am I doing wrong?

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    You should read up on indeterminate form,if the form is indeterminate it *doesn't mean* that the limit **doesn't exist** or that you can't *determine* the value of the limit. – 2017-02-26

4 Answers 4

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Indeterminate forms can have values.

Note from L'Hospital's Rule that $\lim_{n\to \infty}\frac{\log(n)}{n}=\lim_{n\to \infty}\frac{1/n}{1}=0$. Hence, we have

$$\begin{align} \lim_{n\to \infty}\frac{1}{n^{1/n}}&=\lim_{n\to \infty}e^{-\frac1n \log(n)}\\\\ &e^{-\lim_{n\to \infty}\left(\frac1n \log(n)\right)}\\\\ &=e^0\\\\ &=1 \end{align}$$

as expected!

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    Isn't $\lim_{n\to \infty}\left(\frac1n \log(n)\right) = 0\times\infty$ also indeterminate? – 2017-02-26
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    @razzak It is indeterminate. And I showed, using L'Hospital's Rule, that the limit is $0$. – 2017-02-26
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    sorry I got it now, we use L'hopital rule and it ends up as 0. – 2017-02-26
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    @razzak Well done! – 2017-02-26
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Hint :

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

Proof :How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?

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    Have you not simply written the answer that is in question? – 2017-02-26
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$$ \frac{1}{\sqrt[n]{n}}=\left(\frac{1}{n}\right)^{\frac{1}{n}}=e^{\frac{1}{n}\log{(1/n)}}=e^{-\frac{1}{n}\log{n}} $$ since $$ \lim_{n\to\infty}\frac{1}{n}\log{n}=0\implies\lim_{n\to\infty}e^{-\frac{1}{n}\log{n}}=e^0=1 $$

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    you were faster... your answer was not written when i started posting – 2017-02-26
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    No worry; that happens all the time to me too. (+1) for you answer! – 2017-02-26
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It is true that $\infty^0$ is an indeterminate form. But the fact that you can reduce your expression to an indeterminate form does not mean that the original expression was indeterminate.

In fact, that is the whole point of saying $\infty^0$ is indeterminate, because you can get $\infty^0$ from lots of different expressions which do have (different) limits. Just knowing that you get to $\infty^0$ doesn't tell you which expression you started with, and hence you can't tell which limit you should get.

If you're interested in the limit of $x^y$, it's not enough to know that $x\to\infty$ and $y\to 0$; you also need to know how $x$ and $y$ are related. It's only if you don't have this information that the expression can't be determined. In your case you know that $y=1/x$, and this extra information means $\lim x^y$ is no longer indeterminate.

$n^{1/n}>1$, but for any $c>1$ we have $c^n>n$ if $n$ is sufficiently large, so eventually $n^{1/n}