Proof of the derivative of $\sin{x}$ is $\cos{x}$

Question

I recently learned the proof that the derivative of $\sin x$ is $\cos x$ in Stewarts calculus book. However, in his proof he uses preconceived limit laws such as the sum and product law to evaluate the limit. My confusion is that these limit laws can only be used when the limit exists however we do not know the limit in this case we are trying to evaluate it so how can we use these limit laws must we resort to a geometric proof?

Answer 1

There should not be any problem with limits... $$\begin{align}\lim_{h\to 0} \frac{\sin(x + h) - \sin x}{h} &= \lim_{h\to 0} \frac{2\cos(x + \frac h2)\sin\frac h2}{h} \\&= \lim_{h\to 0} \left[\cos\left(x + \frac h2\right)\cdot\frac{\sin (h/2)}{h/2}\right] \\&\overset*= \lim_{h\to 0}\cos\left(x + \frac h2\right)\cdot\lim_{h\to 0}\frac{\sin (h/2)}{h/2} \\&= \cos x \cdot 1 \\&= \cos x\end{align}$$ Explanation for (*): Limit of product is equal to product of limits in this case since both expressions have limits

Answer 2

Let $$f(x)=\sin(x)$$ and $ a,h \in \Bbb R$.

$$A(h)=\frac{f(a+h)-f(a)}{h}=$$ $$\frac{\sin(a+h)-\sin(a)}{h}=$$

$$\frac{\sin(a)\cos(h)+\cos(a)\sin(h)-\sin(a)}{h}=$$ $$\frac{\cos(h)-1}{h}\sin(a)+\frac{\sin(h)}{h}\cos(a)$$ and, $$\lim_{h\to 0}A(h)=0+1.\cos(a)=f'(a)$$