Write a Limit to calculate $f'(0)$ Let $f(x) = \frac {2}{1+x^2} $ 
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is 
$f'(x)$= lim           (f(x+dx)-f(x))/dx
    (dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.

$f'(0) = \displaystyle \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \left (\dfrac{1}{h} \right ) \left ( \dfrac{2}{1 + h^2} - 2 \right )$
$= \lim_{h \to 0}  \left (\dfrac{1}{h} \right )  \left ( -\dfrac{2h^2}{1 + h^2} \right ) = \lim_{h \to 0} -\dfrac{2h}{1 + h^2} = 0. \tag 1$
