What is wrong with this proof that $\sin’x=0$?

Question

I was writing out the proof of $\sin’x=\cos x$ and ended up with something that was wrong, but I’m not sure why. Here it is:

Answer 1

You broke up the initial limit incorrectly. The correct proof would be like this. (Notice how the limit is split up.) $$\sin’x = \lim_{h\to 0}\big(\frac{\sin(x+h)-sin(x)}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x\cos h- \sin x+\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x(\cos h-1)+\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x(\cos h-1)}{h}+\frac{\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\frac{\sin x(\cos h-1)}{h}+\lim_{h\to 0}\frac{\cos x\sin h}{h}$$ $$=\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}$$ $$=\sin x\cdot 0 + \cos x\cdot 1 = \cos x$$ $$\implies \boxed{\sin’x = \cos x}$$

Answer 2

In the second to last line, both limits don't exist. Akin to $\infty-\infty \neq 0$

Answer 3

You have no reason to simplify

$$\frac{\sin x\cos h+\cos x\sin h}{h}$$ in $$\frac{\sin x}h.$$