Really stuck on this one....
$\displaystyle f(x) = \frac{x - \sin{x}}{x^{2}}$ for $x \neq 0$ and $0$ when $x = 0$
Using the definition of the derivative, find $f'(0)$
I know the definition is $$ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
The way I did it was to say $$\lim_{h\to 0}\frac{\frac{(x+h)-\sin(x+h)}{(x+h)^2} - \frac{x - \sin x}{x^2}}{h}$$ $$=\lim_{h\to 0}\frac{\frac{1-\cos(x+h)}{(x+h)^2} - 2\left(\frac{(x+h) - \sin(x+h)}{(x+h)^3}\right)}{1}$$ (using L'Hôpitals Rule) which is $$ \frac{1-\cos(x)}{x^2} -\frac{2(x-\sin x)}{x^3}.$$ But then we cant use this to find $f'(0)$ because the denominator it 0!!!
Where am I going wrong?