$f(z)=\frac{z}{z-2i}\cos (\frac{z}{3})$
It seems that to find the limit will be hard so
$f(z)=\frac{z}{z-2i}\cos (\frac{z}{3})=\frac{z}{z-2i}\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{3^2n\cdot2n!}=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{3^2n\cdot2n!(z-2i)}$
What to do next?