Problem tranforming $\lim_{x\rightarrow 0} \left(\frac{e^{xi} - e^{-xi}}{2ix} \right)$ First of all, thanks for reading this!
I started out with the following limit:
$\lim\limits_{x\rightarrow 0} \left(\frac{e^{xi} - e^{-xi}}{2ix} \right)$, now to my understanding I can transform as follows:
1) Pull $\frac{1}{2}$ out as a constant:
$$\frac{1}{2}\lim\limits_{x\rightarrow 0} \left(\frac{e^{xi} - e^{-xi}}{ix} \right)$$
2) Partition the fraction in the braces as such:
$$\frac{1}{2}\lim\limits_{x\rightarrow 0} \left(\frac{e^{xi} }{ix}- \frac{ e^{-xi}}{ix} \right)$$
3) Bring the parts to the form of a sum by negating the denominator of the second part:
$$\frac{1}{2}\lim\limits_{x\rightarrow 0} \left(\frac{e^{xi} }{ix}+ \frac{ e^{-xi}}{-ix} \right)$$
However in the provided solution, the following transformation is perfomed:
$$\lim_{x\rightarrow 0} \left(\frac{e^{xi} - e^{-xi}}{2ix} \right) = \frac{1}{2}\lim_{x\rightarrow 0} \left(\frac{e^{xi}-1}{ix} + \frac{e^{-xi}-1}{-ix} \right)$$
Now I cannot get my head around which steps is need to perform, to get to that result.
Thank you in advance for any help.
 A: $$ \left(\frac{e^{xi} - e^{-xi}}{2ix} \right) = \frac{1}{2}\left( \frac{e^{xi}-1-e^{-xi}+1}{ix} \right)= \frac{1}{2}\left( \frac{(e^{xi}-1)-(e^{-xi}-1)}{ix} \right)$$
Now do step 2 and 3 you described in your question.
A: Writing as you did brings the limit in a form $\infty-\infty$ (over the imaginary axis), which is much worse than the form $0/0$ you started with.
If you want to separate $e^{xi}$ and $e^{-xi}$, notice that both have limit $1$ for $x\to0$, so the idea is to subtract and add $1$:
$$
\frac{e^{xi}-e^{-xi}}{2ix}=
\frac{e^{xi}-1+1-e^{-xi}}{2ix}=
\frac{e^{xi}-1}{2ix}+\frac{e^{-xi}-1}{-2ix}=
\frac{1}{2}\left(\frac{e^{xi}-1}{ix}+\frac{e^{-xi}-1}{-ix}\right)
$$
Now, both summands have as limit the derivative of $e^z$ at $0$, which is $1$.
This is a general trick: suppose you have two functions $f$ and $g$ such that $f(x_0)=a=g(x_0)$. Then you can compute
\begin{align}
\lim_{x\to x_0}\frac{f(x)-g(x)}{x-x_0}
&=
\lim_{x\to x_0}\frac{(f(x)-a)-(g(x)-a)}{x-x_0}\\[4px]
&=
\lim_{x\to x_0}\left(\frac{f(x)-a}{x-x_0}-
  \frac{g(x)-a}{x-x_0}\right)\\[4px]
&=
\lim_{x\to x_0}\left(\frac{f(x)-f(x_0)}{x-x_0}-
  \frac{g(x)-g(x_0)}{x-x_0}\right)\\[6px]
&=f'(x_0)-g'(x_0)
\end{align}
provided the derivatives exist.
Actually, you don't even need this trick: the limit is the derivative at $x_0$ of the function $h(x)=f(x)-g(x)$ at $x_0$.
In your case, setting $h(x)=e^{xi}-e^{-xi}$, you have $h'(x)=ie^{xi}+ie^{-xi}$, so
$$
\lim_{x\to0}\frac{e^{xi}-e^{-xi}}{2ix}=
\frac{1}{2i}\lim_{x\to0}\frac{e^{xi}-e^{-xi}}{x}=
\frac{1}{2i}h'(0)=\frac{1}{2i}(i+i)=1
$$
