# What is $\lim\limits_{x \to 0^{-}}x^{\frac{1}{2x}}$?

What is $\lim\limits_{x \to 0^{-}}x^{\frac{1}{2x}}$?

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You've tried taking logarithms already? – J. M. May 6 '11 at 16:49
I don't think it is even defined since $x \rightarrow 0^-$ – user17762 May 6 '11 at 16:52
@J.M: I have tried taking logarithms, but encountered the problem Sivaram mentioned above. – Jichao May 6 '11 at 17:30
Yes, and since you're not supposed to have anything negative within the logarithm, then... ? – J. M. May 6 '11 at 17:33

Defining a general exponentiation of $a^b$ for $a,b$ real number is troublesome if you do not require the base ($a$ in this case) to be nonnegative.

This is because we define things for irrational numbers by limits of rational sequences. Now consider a sequence in which the numbers are all such that they have the said root (i.e. odd numbers in the denominator) and another in which there are only even numbers - you have two sequences one whose limit is well defined and another whose values are not even real at any point.

For that reason in real analysis we only define $x^y$ for non-negative $x$ (and we still have to argue the case of $0^0$ to be either $0,1$ or undefined).

So approaching zero from the negative side has little to no meaning in this context, as the function is not well-defined for the valued through which you want to approach the said limit.

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You mean $a$, and later $x$, should be positive, right? There's no reason for the exponents to be positive. – Jonas Meyer May 6 '11 at 18:50
@Jonas: Of course, if the base is positive then all roots are well defined. However if the base is negative then the value of the exponent is only defined when the root is of an odd order, and then - as I said, we have a problem of continuity. – Asaf Karagila May 6 '11 at 18:59

To make the problem well defined, you could use complex logarithms, where you define $\log(re^{i\theta}) = \ln r + i\theta$, with $\theta$ stipulated to be in some range, say $-\pi < \theta \leq \pi$. In this situation, you can define ${\displaystyle x^{1 \over 2x} = e^{\log x \over 2x}}$. Since you are assuming $x < 0$, one has $\log(x) = \ln|x| + i\pi$. So you seek ${\displaystyle \lim_{x \rightarrow 0^-} e^{{\ln|x| + i\pi \over 2x}}}$, which in turn is the same as $$\lim_{x \rightarrow 0^+} e^{-{\ln x + i\pi \over 2x}}$$ $$= \lim_{x \rightarrow 0^+} e^{-{\ln x \over 2x}}e^{-{i\pi \over 2x}}$$ $$= \lim_{x \rightarrow 0^+} x^{-{1 \over 2x}}e^{-{i\pi \over 2x}}$$ By L'hopital's rule, $x^{-{1 \over 2x}}$ goes to $1$ as $x$ approaches zero from the right. On the other hand, $e^{-{i\pi \over 2x}} = \cos({\pi \over 2x}) - i\sin({\pi \over 2x})$ diverges, as it oscillates faster and faster as $x$ approaches zero. So the overall limit does not exist here.

Taking a different branch of the logarithm will lead to the same situation.

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