Derivative of sin(x)/x at $0$ by definition of derivative the question I am attempting is:
Show $f '(0) = 0$ for:
$$f(x) = \left\{
\begin{array}{lr}
\frac{\sin(x)}{x} & : x \neq 0\\
1 & : x=0
\end{array} 
\right.$$
So I got stuck after the following working:
$$
f'(0)  =  \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\
 = \lim_{h \to 0} \frac{\frac{\sin(h)}{h}-\frac{\sin(0)}{0}}{h}\\
= \lim_{h \to 0} \frac{\sin(h)}{h^2}
$$
and the above limit does not exist. Am I not applying the derivative definition right? or am I going wrong somewhere else? 
EDIT: 
I now see that f(0) = 1 from the definition of the function, however I still have:
$$\lim_{h \to 0}\frac{\sin(h)-h}{h^2}$$
which I am struggling to evaluate. Splitting into two fractions the first fraction is still the one from above which does not exist.
Also this must be from definition of derivative directly and not by L'Hopitals rule as it has not been taught yet.

Note that in the usual proof of the limit $\lim_{h\to 0} \frac{\sin(h)}{h} = 1$, it is shown that the inequality $$\cos(h) \le  \frac{\sin(h)}{h} \le 1$$ holds for all $h$ in an interval around $0$.  Assuming this result, we have $$\cos(h) - 1 \le \frac{\sin{h}}{h} - 1 \le 0$$ in this interval.  Taking absolute value and multiplying by $|1/h|$ then yields 
$$0 \le \left|\frac{\sin{h}-h}{h^2}\right| \le \left|\frac{\cos(h)-1}{h}\right|$$
for all $h$ in an interval around $0$.
The desired limit is then $0$ by the squeeze theorem and the standard trig limit $\lim_{x\to 0} \frac{\cos(x)-1}{x} = 0$.
