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