Derivative of $\sin(x)$ issues $$\frac{d}{dx} \sin(x) = \lim_{h\to 0} \frac{\sin(x+h) - \sin(x)}{h}$$
$$\frac{d}{dx} \sin(x) = \lim_{h\to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$
$$\frac{d}{dx} \sin(x) = \sin(x)\lim_{h\to 0} \frac{\cos(h) -1}{h} + \cos(x)\lim_{h\to 0} \frac{\sin(h)}{h}$$
Normally I'd use L'Hopital's Rule here but considering that I'm trying to find the derivative in the first place, that kind of defeats the purpose.
Is there an easier way to approach these limits? I'm not seeing anything obvious.
 A: Better to use this
$$\sin(x+h)-\sin x=2\sin(h/2)\cos(h/2+x)$$
A: Hint: You can use the squeeze theorem to show that
$$\lim_{h\to0}\frac{\sin(h)}{h} = 1.$$
You can also show that
$$\lim_{h\to0}\frac{\cos(h)-1}{h} = 0.$$
Try multiplying top and bottom by $\cos(h) + 1$. You will need to use the first limit.
A: There is no algebraic approach to it. The two I know are:


*

*use some geometric considerations on the unit circle to obtain the two limits by squeezing with obvious limits. 

*Define sine and cosine using their Taylor series. The derivatives are then obvious. 
A: You can try this:
$$\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}=\lim_{h\rightarrow0}\frac{\cos^2(\frac{h}{2})-\sin^2(\frac{h}{2})-\cos^2(\frac{h}{2})-\sin^2(\frac{h}{2})}{h}$$
$$=\lim_{h\rightarrow0}(-\frac{h}{2}*\frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2})=0$$
A: Before studying derivatives you are supposed to be somewhat familiar with limits (and continuity) and when you study these topics you will encounter the following theorem (with or without proof) $$\lim_{x\to 0}\frac{\sin x} {x} =1\tag{1}$$ The typical proof proceeds using the inequalities $$\sin x<x<\tan x$$ for $0<x<\pi/2$. An easy consequence of this is the limit $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{2}$$ and you should be able to deduce $(2)$ from $(1)$ easily using trigonometric identities. Now use the above results to finish your derivation. 
