Does $\lim\limits_{z \to 0}\frac{e^{z^{-1}}}{\sin(z^{-1})}$ exist or not?

Question

I used limit of the function at zero, and got that the limit is zero. So I said, while the limit existed and it is finite then the singularity is Removable Singularity. My function is $$f(z)=\frac{e^{z^{-1}}}{\sin(z^{-1})}$$ I think that I had something wrong while calculating the limit and it should not be existed. Does the limit exist or not? $$\lim_{z \to 0}f(z)=\lim_{z\to0}\frac{e^{z^{-1}}}{\sin(z^{-1})}=?$$

Answer 1

Hint: if the limit exists, then so does the real limit $$\lim_{x \to 0^+} \frac{e^{x^{-1}}}{\sin(x^{-1})} = \lim_{t \to \infty} \frac{e^t}{\sin t}.$$

Answer 2

Assume that $\lim\limits_{z \to 0}\frac{e^{z^{-1}}}{\sin(z^{-1})}$ exists.

Then $\lim\limits_{t \to \infty}\frac{e^{t}}{\sin(t)}$ also exists. But this latter limit does not exist, since $f(t)=\frac{e^{t}}{\sin(t)}$ can take arbitrary large and arbitrary large negative values when $t$ approaches $\infty$.

More precisely, if $x_n=\frac{\pi}{2}+2n\pi$ and $y_n=-\frac{\pi}{2}+2n\pi$, then

$$\lim\limits_{n\to\infty}f(x_n)=\infty\textrm{ and }\lim\limits_{n\to\infty}f(y_n)=-\infty$$

We have a contradiction, so $\lim\limits_{z \to 0}\frac{e^{z^{-1}}}{\sin(z^{-1})}$ cannot exist.