2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – kingW3 Feb 26 '17 at 18:13
3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Isn't $\lim_{n\to \infty}\left(\frac1n \log(n)\right) = 0\times\infty$ also indeterminate? $\endgroup$ – razzak Feb 26 '17 at 18:16
  • $\begingroup$ @razzak It is indeterminate. And I showed, using L'Hospital's Rule, that the limit is $0$. $\endgroup$ – Mark Viola Feb 26 '17 at 18:17
  • $\begingroup$ sorry I got it now, we use L'hopital rule and it ends up as 0. $\endgroup$ – razzak Feb 26 '17 at 18:22
  • $\begingroup$ @razzak Well done! $\endgroup$ – Mark Viola Feb 26 '17 at 18:36
1
$\begingroup$

$$ \frac{1}{\sqrt[n]{n}}=\left(\frac{1}{n}\right)^{\frac{1}{n}}=e^{\frac{1}{n}\log{(1/n)}}=e^{-\frac{1}{n}\log{n}} $$ since $$ \lim_{n\to\infty}\frac{1}{n}\log{n}=0\implies\lim_{n\to\infty}e^{-\frac{1}{n}\log{n}}=e^0=1 $$

$\endgroup$
  • $\begingroup$ you were faster... your answer was not written when i started posting $\endgroup$ – Francesco Alem. Feb 26 '17 at 18:12
  • $\begingroup$ No worry; that happens all the time to me too. (+1) for you answer! $\endgroup$ – Mark Viola Feb 26 '17 at 18:17
1
$\begingroup$

Hint :

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

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

$\endgroup$
  • $\begingroup$ Have you not simply written the answer that is in question? $\endgroup$ – Mark Viola Feb 26 '17 at 18:08
0
$\begingroup$

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 $n^{1/n}<c$. This means that the limit is $1$ in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.