Find the limit of sequence (complex numbers)

Question

How do I find the limit of $$Z_n=n\sin{\frac{i}{n}}$$ I'd say it is 0, but the book says $i$.

Answer 1

$Z_n= i\frac {\sin (i/n)} {i/n} \to i$.

Answer 2

Hints:

$$\sin(ix) = i\cdot \sinh b\\\lim_{x\to0} \frac{\sinh(x)}{x} = 1$$