(i) The Derivative of the Sine at $x=0$: $\sin'(0)=1$.
Proof:

Using the definition of the derivative at a number,

we have the  derivative of $\sin$ at $0$, denoted $\sin'(0)$, is defined as  $$\sin'(0) = \lim_{x \rightarrow 0} \frac {\sin(x)-\sin(0)}{x-0}$$ or  $$\sin'(0) = \lim_{h \rightarrow 0} \frac {\sin(0+h)-\sin(0)}{h} = \lim_{h \rightarrow 0} \frac {\sin(h)}{h}.$$

From work with the trigonometric function related to the circle it can be shown that for $h \gt 0$,

$\cos(h) < \frac {\sin(h)} h < 1$.
This inequality shows that when $h$ is close to $0$, $h>0$, $\frac {\sin(h)} h$ is between $\cos(h)$ and $1$.

Furthermore, when $h$ is close to $0$, $\cos(h)$ is close to $1$. These facts are illustrated with mapping diagrams to visualize $\cos(h) < \frac {\sin(h)} h < 1$.
Since $\frac {\sin(h)} h$ must be even closer to $1$ than $\cos(h)$, if $h > 0$ and $h \to 0$, we see that $\frac {\sin(h)} h \to 1$. That is,

$$\lim_{h \to 0^+} \frac {\sin(h)} h = 1$$.

Now what about the case when $h < 0$ and $h \to 0$?

In this case we have $\frac {\sin(h)} h = \frac {\sin(-h)}{-h}$. To see why this is true, let $k = -h$. Recall that the sine functions is odd, so $\sin(-h) = -\sin(h)$, so $\sin(k) = -\sin(h)$. When $h<0, k = -h > 0$, and if $h \to 0$, then $k = -h \to 0$ as well. Thus we see that
$\frac {\sin(h)} h = \frac {-\sin(h)}{-h} = \frac {sin(k)} k$.

Our previous analysis  showed that when $k > 0$ and $k\to 0$,  $\frac {\sin(k)} k \to 1$.
So when $h < 0$ and $h \to 0$, $\frac {\sin(h)} h= \frac {\sin(k)} k \to 1$, i.e.
$$\lim_{h \to 0^-} \frac {\sin(h)} h = 1$$.
To summarize, in either case, with $h>0$ or with $h<0$, we find that when $h\to 0$, $\frac {\sin(h)} h \to 1$.  Therefore
$$\sin'(0) = \lim_{h \to 0} \frac {\sin(h)} h = 1.$$   EOP