More limits and derivatives, can I do this? Here is the given problem:
Suppose $f(0)=0$ and $f'(0)=-1$. Evaluate the following limit if it exists:
$$\lim\limits_{h \to 0} \frac{f(h)-f(2h)+f(3h)-f(4h)+...+f(2013h)}{h}$$
So what I was thinking was to split it up so you have,
$$\lim\limits_{h \to 0} \frac{f(h)}{h} + \lim\limits_{h \to 0} \frac{f(3h)-f(2h)}{h}+...+\lim\limits_{h \to 0} \frac{f(2013h)-f(2012h)}{h}$$
So the first part is equal to $0$ and the rest is MVT and you get 
$$f'(c_1)+...+f'(c_{1006})$$
which each would equal $-1$ and total $-1006$. Is this correct?
 A: Let $\lambda$ be a constant. Then
$$ \lim_{h\to 0} {f(\lambda h)\over h} = \lambda \lim_{h\to 0} {f(h)\over h} = \lambda f'(0) = -\lambda.$$
Now sum it up.  Notice this still works if $\lambda = 0$; you see this separately since the division done here would be illegal in that case.
A: This is much easier than it looks. Notice that if you let $h=0$ in the first limit, you obtain $0/0$. This means that you can apply l'Hopital's rule and make this easier on yourself. Differentiating the top and bottom with respect to $h$ yields
$$
\lim_{h \to 0} \frac{f'(h)-2f'(2h)+3f'(3h)-\cdots+2013f'(2013h)}{1}
$$
Letting $h \to 0$ and using the fact that $f'(0)=-1$, you have 
$$
-1+2-3+4-\cdots-2013
$$
which is $-1007$--as in nemathsadist's answer. 
A: You solved it, but your own solution has a little problem.
You considered $\lim\limits_{h \to 0} \frac{f(h)}{h}$ to be zero, but you must know that the expression is in $\frac00$ form, so to evaluate that part, you have to use L'Hopital rule or any other rule you wish to use to find its value.
Applying L'Hopital, we get, $\lim\limits_{h \to 0} \frac{f'(h)}{1}$ or $\lim\limits_{h \to 0} \frac{f'(0)}{1}$. Its value is -1.
A: ncmathsadist has already given a perfect answer. I want to highlight the issue with your approach. A derivative $f'(0) = -1$ means that $$\lim_{h \to 0}\frac{f(h) - f(0)}{h} = -1$$ It does not mean that $$\lim_{h \to 0}\frac{f(3h) - f(2h)}{h} = -1$$ (in fact this limit might not even exist). Hence your split does not work. Note the way ncmathsadist uses $f(0) = 0$ in each term and that leads to the derivative so that the final answer is $-1+2-3 + \cdots - 2013$
