Find the residue of $$f(z)=\sin(1/z)$$ in its singularity point.
My attempt: I have 2 different answers so my question is: where is my mistake in one of the answers?
Answer 1: $$\nexists \lim_{z\to 0}f(z) \\ \lim_{z\to0}f(z)\cdot z=\lim_{z\to0}\sin(1/z)/(1/z)$$ But $$ |{\sin(1/z)\over 1/z}|\leq|{1\over (1/z)}|=|z|\to0 $$ Thus $$\lim_{z\to 0}f(z)\cdot z=0$$Then $z=0$ is a simple pole and $\operatorname{Res}(f,0)=0$.
Answer 2:
In the Laurent expansion of $f$ around $z=0$, the coefficient $c_{-1}$ equals $1$, thus the residue equals $1$.
The first answer is wrong. It is false that$$(\forall z\in\mathbb C\setminus\{0\}):\left\lvert\sin\left(\frac1z\right)\right\rvert\leqslant1.$$In fact$$\lim_{t\to0^+}\left\lvert\sin\left(\frac it\right)\right\rvert=\infty.$$