Why does $\lim_{x\to0^{-}} \mathrm {Im}\left( \mathrm \ln \left(x\right)e^x\right)=\pi$?

Question

Why does $$\lim_{x\to 0^{-}} \mathrm {Im} \left( \ln\left(x\right) e^x\right)=\pi$$ Obviously this is no coincidence. I was thinking maybe this has to do with Euler's formula, but I don't see how the logarithm factors in. Does this even have to do with Euler's Formula?

Answer 1

For values of $x$ along the negative real axis, we can write $x=|x|e^{i(2\ell+1)\pi}$ for integer values of $\ell$. Therefore, the multi-valued logarithm is given by

$$\log x=\log (|x|e^{i(2\ell +1)\pi})=\log |x|+i (2\ell+1) \pi)$$

Finally,

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^{-}} \text{Im}\left(\log(x) e^{x}\right)=(2\ell+1) \pi}$$

If we restrict ourselves to the principal branch of the complex logarithm, then

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^{-}} \text{Im}\left(\log(x) e^{x}\right)=\pi}$$

Answer 2

$$\lim_{x \to 0^-}\ln x e^x \stackrel{e^x \to 1}{=}\lim_{x \to 0^-}\ln x = \lim_{x \to 0^+}\ln (-x) = \ln(-1) + \lim_{x \to 0^+}\ln (x) $$

The real part of the final expression is undefined ($\lim_{x \to 0^+}\ln (x) = -\infty$), and the imaginary value depends on the branch under consideration. The imaginary part of the principal value is $\pi$ (because the principal value of $\ln(-1) = i\pi$).