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.
 A: 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 $\lim_{x\to 0} \frac{\cos(x)-1}{x} = 0$.
A: Your issue is in evaluating $f(0)$ as $(\sin{0})/0$: you haven't used the definition you have given, which has $f(0)=1$.
Then you have
$$ \lim_{h \to 0} \frac{\frac{\sin{h}}{h}-1}{h} = \lim_{h \to 0} \frac{\sin{h}-h}{h^2} = \lim_{h\to 0} \frac{\cos{h}-1}{2h} = \lim_{h \to 0} \frac{-\sin{h}}{2} = 0, $$
using L'Hôpital's rule twice.

To avoid using L'Hôpital, the easiest way is to use the squeeze theorem: for $0<x<\pi/2$,
$$ \sin{x}<x<\tan{x}, \tag{1} $$
which can be shown by drawing triangles inside and tangent to the unit circle. Dividing by $\sin{x}$ gives
$$ 1<\frac{x}{\sin{x}}<\frac{1}{\cos{x}}, $$
which shows, taking $x$ to $0$, that the limit of $x/\sin{x}$ must be $1$. The difficult bit is now to show that the limit of $\frac{\sin{h}-h}{h^2}$ is actually zero.
The inequality (1) also shows that
$$ 0<1-\frac{\sin{h}}{h}<\frac{\tan{h}-\sin{h}}{h} = \frac{\sin{h}}{h}\frac{1-\cos{h}}{\cos{h}} $$
Using the identity $1-\cos{h}=2\sin^2{(h/2)}$, and dividing by $h$, we have
$$ 0<\frac{h-\sin{h}}{h^2} < \frac{\sin{h}}{h} \frac{2\sin{(h/2)}}{h} \frac{1}{\cos{h}} (2\sin{(h/2)})  $$
We know enough limits to compute the right-hand side: the first 3 factors all tend to $1$, and the last one tends to zero, so another sandwich theorem application gives
$$ \lim_{h \to 0} \frac{\sin{h}-h}{h^2} =0. $$
I'm quite surprised that this can actually be pushed this far.
A: In this answer, it is shown that for $0\le x\le\frac\pi2$,
$$
\cos(x)\le\frac{\sin(x)}x\le1\tag{1}
$$
So, for $0\lt x\le\frac\pi2$, and symmetrically for $-\frac\pi2\le x\lt0$, subtract $1$ from $(1)$ and divide by $x$:
$$
0=\lim_{x\to0}\frac{\cos(x)-1}{x}\le\lim_{x\to0}\frac{\frac{\sin(x)}x-1}{x-0}\le0\tag{2}
$$
We have the leftmost equality because
$$
\begin{align}
\lim_{x\to0}\frac{1-\cos(x)}x
&=\lim_{x\to0}\frac1x\frac{\sin^2(x)}{1+\cos(x)}\\
&=\lim_{x\to0}\frac{\sin(x)}x\lim_{x\to0}\frac{\sin(x)}{1+\cos(x)}\\[6pt]
&=1\cdot0\tag{3}
\end{align}
$$
Therefore, by the Squeeze Theorem, $(2)$ says
$$
f'(0)=0\tag{4}
$$
