Prove that this limit exist

Question

\begin{aligned} \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h} \end{aligned}

I tried to find the $\delta$ and $\epsilon$. My attempts to limit has been in vain and I can not clear the $h$.

tried: \begin{aligned} \left|\frac{\sin(x+h)-\sin(x)}{h}\right|<\varepsilon \newline \left|\frac{\sin(x)\cdot\cos(h)+\sin(h)\cdot\cos(x) - \sin(x)}{h}\right|<\varepsilon \newline \left|\frac{\sin x\cdot(\cos(h)-1)}{h} + \frac{\cos x(\sin(h)-h)}{h}\right| < \varepsilon \newline \end{aligned} applying triangular inequality and delimiting
\begin{aligned} 0<\sin x < 1 \end{aligned} \begin{aligned} 0<\cos x<1 \end{aligned} ... I can not find my $| h | < \delta$.

Answer 1

Note that $$\sin (x+h)=\sin x \cos h + \cos x \sin h$$ and also $$\cos h=1-2\sin^2 \frac h2$$ so that $$sin (x+h)-\sin x= -2\sin x \sin^2 \frac h2 +\cos x\sin h$$

Now you need a standard result for $\lim_{h\to 0}\frac {\sin h}h$

Answer 2

Recall that (take a look to this link):

$$\sin\alpha - \sin\beta = 2 \cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}.$$

Therefore:

$$\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h} = \lim_{h\to0} \frac{2}{h} \cos\frac{2x+h}{2}\sin\frac{h}{2}.$$

Since

$$\lim_{t \to 0} \frac{\sin(t)}{t} = 1,$$

then, the previous limit reduces to:

$$\lim_{h\to0} \cos\frac{2x+h}{2} = \cos(x).$$