Derivative definition Hey I have 2 derivative definition which were told in class.
First one is 
$ \underset{x\rightarrow a}{\lim}\frac{f(x)-f(a)}{x-a} = f'(a) $  This on is pretty straight forward for me.
The second one is 
$ \underset{h\rightarrow0}{\lim}\frac{f(x_{0}+h)-f(x_{0}-h)}{2h} = f'(x_0)$ this one however isn't clear enough. 
The explanation involved something like $\frac{f(x_{0}+h)-f(x_{0}-h)}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}+\frac{f(x_{0}-h)-f(x_{0})}{-h};$
And $\underset{h\rightarrow0}{\lim}\frac{f(x_{0}+h)-f(x_{0})}{h}=\underset{h\rightarrow0}{\lim}\frac{f(x_{0}-h)-f(x_{0})}{-h}=f'(x_{0});$
 (this one is pretty clear to me it's like to substitute $h=x-x_{0}$
Then $\frac{f(x_{0}+h)-f(x_{0}-h)}{2h}=\frac{1}{2}\left(\frac{f(x_{0}+h)-f(x_{0})}{h}+\frac{f(x_{0}-h)-f(x_{0})}{-h}\right)=\frac{1}{2}(2f'(x_{0})=f'(x_{0})$
Now a question in my homework is to express the following expression in terms of $f'(x)$ 
The expression is :
$$\underset{h\rightarrow0}{\lim}\frac{f(x_{0}-2h)-f(x_{0}+3h)}{h}$$
Then I try work the expression and get $\frac{f(x_{0}-2h)-f(x_{0}+3h)}{h}=\frac{f(x_{0}-2h)-f(x_{0})}{h}+\frac{f(x_{0}+3h)-f(x_{0})}{-h}$
Now, is it correct that  $\underset{h\rightarrow0}{\lim}\frac{f(x_{0}-2h)-f(x_{0})}{h}=\underset{h\rightarrow0}{\lim}\frac{f(x_{0}+3h)-f(x_{0})}{-h}=f'(x)
 $   ?
if yes my expression is simply  $2f'(x_0)$  however I am not sure and will appreciate some explanation.
 A: Note that the definition of derivative of a function $f$ defined in a certain neighborhood of $x_{0}$ is given by $$f'(x_{0}) = \lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}} = \lim_{h \to 0}\frac{f(x_{0} + h) - f(x_{0})}{h}\tag{1}$$ The thing to note here is that the derivative of $f$ at $x_{0}$ crucially depends on the value of $f$ at $x_{0}$ i.e. $f(x_{0})$ as well as the values of $f$ near $x_{0}$ (i.e. $f(x_{0} + h)$). So if you want to link any limit with some sort of derivative then you need to identity the point $x_{0}$ and also need to express your limit exactly in the form $(1)$ above.
The second definition is not a definition of derivative, but rather it is a simple theorem:
A Simple Theorem: If $f'(x_{0})$ exists then $$\lim_{h \to 0}\frac{f(x_{0} + h) - f(x_{0} - h)}{2h} = f'(x_{0})$$
The proof is as given in your question i.e. you have to bring the $f(x_{0})$ somehow into picture before you can link it with derivative $f'(x_{0})$ and this is done by writing $f(x_{0} + h) - f(x_{0} - h)$ as $$f(x_{0} + h) - f(x_{0}) - (f(x_{0} - h) - f(x_{0}))$$ Note that the above result does not say anything about the scenario when $f'(x_{0})$ does not exist.
Next you are asked to calculate $$\lim_{h \to 0}\frac{f(x_{0} - 2h) - f(x_{0} + 3h)}{h}$$ The only way now is to introduce $f(x_{0})$ also into the picture and we can proceed as follows:
\begin{align}
L &= \lim_{h \to 0}\frac{f(x_{0} - 2h) - f(x_{0} + 3h)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x_{0} - 2h) - f(x_{0}) + f(x_{0}) - f(x_{0} + 3h)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x_{0} - 2h) - f(x_{0})}{h} - \frac{f(x_{0} + 3h) - f(x_{0})}{h}\notag\\
&= \lim_{h \to 0}-2\cdot \frac{f(x_{0} - 2h) - f(x_{0})}{-2h} - 3\cdot\frac{f(x_{0} + 3h) - f(x_{0})}{3h}\notag\\
&= -2f'(x_{0}) - 3f'(x_{0})\notag\\
&= -5f'(x_{0})\notag
\end{align}
It is important to remember that if you wish to link any limit with $f'(x_{0})$ then it is essential to first get $f(x_{0})$ and then transform the expression such that some part of it looks like the expression in equation $(1)$.
Update: A more common blunder in such questions is the use of L'Hospital's rule to reduce the fraction $$\frac{f(x_{0} - 2h) - f(x_{0} + 3h)}{h}\tag{2}$$ to $$\frac{-2f'(x_{0} - 2h) - 3f'(x_{0} + 3h)}{1}\tag{3}$$ and then taking limit as $h \to 0$. There are two problems in this approach. First L'Hospital's Rule is a complicated thing and many beginners don't know the exact conditions under which it can be applied and conditions under which it is useful. In the current case we don't know if the function $f$ is differentiable at points near $x_{0}$ like $x_{0} - 2h$ and $x_{0} + 3h$. We are only given that $f'(x_{0})$ exists. Next problem is taking limit of $(3)$ as $h \to 0$ after L'Hospital Rule. This limit will exist only when $f'(x)$ is continuous at $x = x_{0}$. This is also not given in question.
A: It is wrong. It should be (for example)
$$\underset{h\rightarrow0}{\lim}\frac{f(x_{0}-2h)-f(x_{0})}{h}
=-2\underset{h\rightarrow0}{\lim}\frac{f(x_{0}-2h)-f(x_{0})}{-2h}
=-2f'(x_0).$$
The other is similar, which can be left as an exercise.
A: To make it as clear as possible, we can separate this expression in a sum of two parts:
$$\frac{f(x_{0}-2h)-f(x_{0}+3h)}{h}=\frac{f(x_{0}-2h)-f(x_{0}+2h)}{h}+\frac{f(x_{0}+2h)-f(x_{0}+3h)}{h}$$
$$\lim_{h \to 0} \frac{f(x_{0}-2h)-f(x_{0}+2h)}{h}=4\lim_{2h \to 0} \frac{f(x_{0}-2h)-f(x_{0}+2h)}{4h}=-4f'(x_0)$$
(since $2h+2h=4h$)
$$\lim_{h \to 0} \frac{f(x_{0}+2h)-f(x_{0}+3h)}{h}=-f'(x_0)$$
(since $3h-2h=h$)
Finally:
$$\lim_{h \to 0} \frac{f(x_{0}-2h)-f(x_{0}+3h)}{h}=-f'(x_0)-4f'(x_0)=-5f'(x_0)$$

Of course, it makes as much sense to just use: $3h+2h=5h$
Note, that even though this 'derivative' is not symmetric, the displacement is proportional to $h$, so it vanishes in the limit $h \to 0$
To illustrate change the variable $x_0 \to x_0+2h$:
$$\lim_{h \to 0} \frac{f(x_{0}+2h+h)-f(x_{0}+2h)}{h}=\lim_{h \to 0} f'(x_0+2h)= f'(x_0)$$
Edit
Apparently, this reasoning is flimsy, but at least the limit is correct - see
http://www.wolframalpha.com/input/?i=lim+h->0+(f(x%2B3h)-f(x%2B2h))%2Fh
