text: Q:
  
  Solving limit of $\frac{1}{x+1}(e^{2\pi i(x+1)}-1)$ as $x\rightarrow -1$ without l'Hopital
  
  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. 
  
  A:
  
  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 $\lim_{t\to 0} \frac{\sin t}{t}$?
  
  