Limit of complex numbers What would be the limit of following term?
$$\lim_{n \to \infty} \frac{e^{inx}}{2^n}$$
I tried to convert the $e^{inx}$ into trigonometric form and tried to do some simplification but got stuck after that. 
Thanks. :)

Hint
Just as you tried $$\frac{e^{inx}}{2^n}=\frac{\cos(nx)+i\sin(nx)}{2^n}$$ Each sine and cosine being between $-1$ and $1$, the numerator is finite and the denominator goes to $\infty$; thus, the limit is $0$.
