Find $\lim_{x→0^-} f(x)$ if $f(x) =x^x$,

Question

$x^x = (e^{\log x})^x = e^{x \log x}$

$\lim_{x\rightarrow0^-} e^{x \log x} = e^{\lim_{x\rightarrow0^-} x \log x}$

$\lim_{x\rightarrow0^-} x \log x = \lim_{x\rightarrow0^-} \frac{\log x}{(\frac{1}{x})} = \lim_{x\rightarrow0^-} \frac{(\log x)'}{(\frac{1}{x})'} = \lim_{x\rightarrow0^-} \frac{\frac{1}{x}}{(-\frac{1}{x^2})} = \lim_{x\rightarrow0^-} -x = 0$

Im stuck

Perfect! So $\lim_{x\rightarrow0^+} x^x = e^0 = 1$. The limit from the left is not defined. — Mark McClure, Oct 14 '14 at 22:50
When you say "$\lim_{x\rightarrow0^-} e^{x \log x} = e^{\lim_{x\rightarrow0^-} x \log x}$", is $\log x$ well defined for negative $x$? — Henry, Oct 14 '14 at 23:00
@Alizter Not quite; that one is about defining $0^0$, not about the limit. I could not find a duplicate for this particular (odd) question. — , Oct 15 '14 at 00:16

score 0 · Answer 1 · answered Oct 14 '14 at 23:15

You will have no luck in finding the limit as $x$ approaches zero from below, because the complex function $$ f(z) = z^z $$ Has a branch cut along the negative real axis ending at the origin.

Thus for example, taking $z \rightarrow 0$ along the imaginary axis, we find that $(iy)^{iy}$ goes to $-\infty$ as $y \rightarrow 0^+$. Whenever there is an endpoint of a branch cut of a function, there will be some direction along which the limit is undefined, and the negative real axis is that path for this function.

While this is a very informative, it is way above the level of the poster and is entirely unhelpful to OP. — Cameron Williams, Oct 14 '14 at 23:31

Find $\lim_{x→0^-} f(x)$ if $f(x) =x^x$,

1 Answers1