I'm doing some proofs establishing the derivatives for some complex functions, and I was asked to show $(\sinh (z))' = \cosh (z)$. Now I know how to do this with the difference quotients ad such, but i am stuck at the step establishing:
$$\lim_{h\to0} \frac{\cosh(h) - 1}{h} = 0$$
Now for all intents and purposes I am supposing this is correct just by the construction of the problem, but in all my years of math I have never had to personally establish that relationship as well as the one for $\sinh(z)/z$, I was wondering if anybody could direct me to where I could find the proofs because they have most likely already been done and are probably tedious.
Thanks in advance
Since $\cosh(h)=\frac{e^h}{2}+\frac{e^{-h}}{2}$, the limit is $$\frac{1}{2}\lim_{h\to 0}\left(\frac{e^h-1}{h}\right)+\frac{1}{2}\lim_{h\to 0}\left(\frac{e^{-h}-1}{h}\right)=\frac{1}{2}\lim_{h\to 0}\left(\frac{e^h-1}{h}\right)-\frac{1}{2}\lim_{h\to 0}\left(\frac{e^{h}-1}{h}\right)=0.$$ Note that we have to use the fact that the limit exists here, but we know it does because $e^x$ is differentiable, and so $\cosh$ is the sum of differentiable functions.