Write a Limit to calculate $f'(0)$

Question

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.

Answer 1

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

Answer 2

it is $$\frac{f(x+h)-f(x)}{h}=\frac{\frac{2}{1+(x+h)^2}-\frac{2}{1+x^2}}{h}$$ Can you finish? Ok, another hint: The numerator is given by $$-2\,{\frac {h \left( h+2\,x \right) }{ \left( {h}^{2}+2\,xh+{x}^{2}+1 \right) \left( {x}^{2}+1 \right) }} $$