Taking the derivative of $\frac1{x} - \frac1{e^x-1}$ using the definition Given $f$:
$$
f(x) = \begin{cases}
        \frac1{x} - \frac1{e^x-1} & \text{if } x \neq 0 \\
        \frac1{2}                 & \text{if } x = 0
\end{cases}
$$
I have to find $f'(0)$ using the definition of derivative (i.e., limits). I already know how to differentiate and stuff, but I still can't figure out how to solve this. I know that I need to begin like this:
$$
f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{\frac1{h} - \frac1{e^h-1}-\frac1{2}}{h}
$$
But I don't know how to do this. I feel like I should, but I can't figure it out. I tried distributing the denominator, I tried l'Hôpital's but I get $0$ as the answer, while according to what my prof gave me (this is homework) it should be $-\frac1{12}$. I really don't know how to deal with these limits; could someone give me a few tips?

Hmm, another approach, which seems simpler to me. However I'm not sure whether it is formally correct, so possibly someone else can also comment on this.     
The key here is that the expression $\small {1 \over e^x-1 } $ is a very well known generation function for the bernoulli-numbers
$$\small {1 \over e^x-1 } = 
x^{-1} - 1/2 + {1 \over 12} x - {1 \over 720} x^3 + {1 \over 30240 }x^5 + O(x^7) $$    
from where we can rewrite
$$\small \frac1x - {1 \over e^x-1 } = 1/2 - {1 \over 12} x + {1 \over 720} x^3 - {1 \over 30240 }x^5 + O(x^7) \qquad \text{ for } x \ne 0 $$
and because by the second definition $\small f(x)=\frac12 \text{ for }x=0$ that power series is the analytic expression for both cases at and near that point.    
Then the derivative can be taken termwise:
$$\small (\frac1x - {1 \over e^x-1 })' = - {1 \over 12}  + {3 \over 720} x^2 - {5 \over 30240 }x^4 + O(x^6)  $$
and is $\small -\frac1{12} $ at x=0 
