# Find $\lim_{x\to 0}x^{\frac{1}{x}}$

I tried to solve $$\lim_{x\to 0} x^{\frac{1}{x}}$$

I tried doing it like this:

$$\lim_{x\to 0} x^{\frac{1}{x}}= \lim_{x\to 0} e^{\ln(x^{\frac{1}{x}})} = \lim_{x\to 0} e^{\frac{\ln(x)}{x}} = \exp\left(\lim_{x\to 0} \frac{\ln(x)}{x}\right)$$

Then solving $\lim_{x\to 0} \frac{\ln(x)}{x}$ with l'Hospital's Rule I get

$$\lim_{x\to 0} \frac{\ln(x)}{x}=\lim_{x\to 0} \frac{1}{x}$$

Which remains indeterminate as it has no two-sided limit.

So is that the answer, that it remains indeterminate? Or would I say that it has two limits depending on the side you approach? How do I phrase this result?

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You'd need to find $\lim\limits_{x\to 0^+} x^{\frac{1}{x}}$ and $\lim\limits_{x\to 0^-} x^{\frac{1}{x}}$. –  user2468 Aug 23 '12 at 5:29
Perhaps even before beginning the calculations you should make clear for yourself what is $\,\displaystyle{x^{1/x}}\,$ for negative arguments....for example, what is $\,\displaystyle {(-1000)^{-1/1000}}=\,$ the $\,1000\,$-th root of $\,-1/1000\,$...?!? –  DonAntonio Aug 23 '12 at 5:33
It is not proper to use L’Hospital for $\lim_{x\to0}(\log x)/x$ because it is not of the form $0/0$ nor $\infty/\infty$. –  Lubin Sep 12 '12 at 19:44

The expression $\lim_{x\to 0^-}x^{\frac{1}{x}}$ makes no sense, for in general $x^y$ is undefined if $x$ is negative. As a consequence, $\lim_{x\to 0}x^{\frac{1}{x}}$ does not exist.

As was pointed out by Lubin, L'Hospital's Rule is not appropriate here, since after taking the logarithm we are looking at $\frac{\ln x}{x}$. For $x$ near $0$, the numerator $\ln x$ is very large negative, but the denominator is close to $0$, so we do not have a suitable form. But luckily, the rule is not needed, since raising a number close to $0$ to a large power gives a number very close to $0$.

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His procedure is not right because L'Hospital does not apply. –  Lubin Sep 13 '12 at 16:49

Don't use L’Hospital, it doesn’t apply. And don’t bother using $\log$ and $\exp$, just plug in a particular tiny number for $x$ and see what happens. For instance, try $x=.1$: then you’re asking about the tenth power of $.1$, which of course is $.0000000001$. Now you see that your argument is to say that you’re raising a tiny number to a huge power, result will be ZERO in the limit.

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$\lim_{x\to 0} x^\frac{1}{x}$=$\lim_{x\to 0} (1+x-1)^{{\frac{1}{x-1}\cdot}{\frac{x-1}{x}}}$=$e^{{\lim_{x\to 0}}{\frac{x-1}{x}}}$=$e$.

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No.    –  Did Sep 12 '12 at 19:14