Solving limit of $\frac{1}{x+1}(e^{2\pi i(x+1)}-1)$ as $x\rightarrow -1$ without l'Hopital

Question

I have no idea how to solve the limit of

$$\frac{1}{x+1}(e^{2\pi i(x+1)}-1)$$ as $x\rightarrow -1$, without using l'Hopital's rule.

I don't even know where to start at all. Any hints would be greatly appreciated.

Answer 1

Hint: $\exp(2\pi i(x+1))-1 = \cos(2\pi(x+1))-1+i\sin(2\pi(x+1))$.

Can you say anything about $\lim_{t\to 0} \frac{\cos t-1}{t}$ and $\lim_{t\to 0} \frac{\sin t}{t}$?

Answer 2

Hint: Use taylor expansion of $\exp(u)=1+u+u^2/2+O(u^3)$, in which you replace $u=2\pi i(x+1)$.

Remember that you have to change the limit.

Answer 3

$$ \lim_{x\to -1} \frac{e^{2\pi i(x+1)}-e^{2\pi i(-1+1)}}{x-(-1)} $$ is by definition the derivative of $e^{2\pi i(x+1)}$ at $x=-1$.

Differentiate that symbolically (chain rule!), plug in $-1$, and you're done.