How to prove $\lim_{x\to \infty}\Im\left(xi^{1/x}\right)=\frac{\pi}{2}$?

Question

I have yet to study complex analysis, but I sperimentally found $$\lim_{x\to \infty}\Im\left(xi^{1/x}\right)=\frac{\pi}{2}.$$ W|A agrees with me too, and while I know that's not so significant, knowing Euler's identity I think it makes sense. How does one prove it?

Answer 1

$$i=\exp(\frac{\pi i}{2})\Rightarrow (i)^{1/x}=(\exp(\frac{\pi i}{2}))^{1/x}=\exp(\frac{\pi i}{2x}+\frac{2\pi ik}{x})=\exp(\frac{(1+4k)\pi i}{2x})$$ Using Euler's formula $$\exp(\frac{(1+4k)\pi i}{2x})=\cos(\frac{(1+4k)\pi}{2x})+i\sin(\frac{(1+4k)\pi}{2x})$$$$\Rightarrow \Im(\exp(\frac{(1+4k)\pi i}{2x}))=\sin(\frac{(1+4k)\pi}{2x})$$ Now $$\lim_{x\to\infty}x\sin(\frac{(1+4k)\pi}{2x})=\frac{(1+4k)\pi}{2}\lim_{x\to\infty}\frac{\sin(\frac{(1+4k)\pi}{2x})}{\frac{(1+4k)\pi}{2x}}=\frac{(1+4k)\pi}{2}$$ where $k\in\mathbb{Z}$.

Answer 2

$$\mathfrak{I} xi^{1/x}=\mathfrak{I} x e^{\pi i/2x}=x\sin \pi/2x$$.

Now use $\sin \delta\sim \delta $ for $\delta$ being small.

Answer 3

$$\lim_{x\to\infty}\Im x i^{1/x} = \lim_{y\to 0}\Im\frac{i^y}{y}=\lim_{y\to 0}\Im\frac{e^{y\log i}}{y}=\lim_{y\to 0}\Im\frac{e^{iy\pi/2}}{y}=\lim_{y\to 0}\frac{\sin(y\pi/2)}{y}=\frac{\pi}{2}\lim_{y\to 0}\frac{\sin y\pi/2}{\pi/2 y}=\frac{\pi}{2}.$$