Thread: Finding derivatives

Check my work please

Find $\displaystyle f'(0)$ if

$\displaystyle f(x)=\frac{k(x)}{x}$ if $\displaystyle x $ not $\displaystyle 0$
$\displaystyle 0 $ if $\displaystyle x=0$

and $\displaystyle k(0)=k'(0)=0$ and $\displaystyle k''(0)=17$

My solution
$\displaystyle f'(0)=lim_{h\rightarrow{0}} \frac{f(h)-f(0)}{h}$
$\displaystyle =lim_{h\rightarrow{0}} \frac{\frac{k(h)}{h}}{h}$
Question boils down to finding $\displaystyle lim_{h\rightarrow{0}} k(h)$

But k is differentiable at 0, hence k is continuous at 0, so
$\displaystyle lim_{h\rightarrow{0}} k(h)=0$
Hence $\displaystyle f'(0)=0$
END

I am concerned about my solution because it doesn't use some of the information provided. Help please?

$\displaystyle \lim \frac{k(h)}{h^2} = \lim \frac{k'(h)}{2h} = \lim \frac{k''(h)}{2} = \frac{k''(0)}{2} = \frac{17}{2}$

I applied L'Hopital's rule twice. All limits are taken as h goes to 0.