Calculate limit of complex function

Question

Why is $$\lim\limits_{n \to \infty} \frac{1}{n} \frac{1-e^{it}}{e^{\frac{-it}{n}} - 1} = \frac{e^{it}-1}{it}?$$

I tried applying L'Hospital but it failed and gives me $e^{it}$ as the limit.

Answer 1

Essentially you can rewrite the limit using $x=1/n$ to get $$ (1-e^{i t})\cdot \lim_{x\to0} \frac{x}{e^{- i t x} - 1} $$ You can apply l'Hopital's rule in the last equation to arrive at the result.

Answer 2

HINT

Use that

$$e^{\frac{-it}{n}} =1-\frac{it}{n}+o\left(\frac1n\right)$$