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

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:

• For a general $x$, for instance for $x=\pi/4$, $$\frac{\sin x}h$$ has no limit for $h\to 0$, so the first split makes no sense. – dan_fulea Oct 1 '18 at 17:31
• How you jumped from 3rd line to 4th? you will have a term of $\cos{x}$, as, $\lim_{h\to 0}\frac{\sin{h}}{h}=1$ – tarit goswami Oct 1 '18 at 17:31
• So $\infty-\infty=0?\ \ddot\frown$ – Lord Shark the Unknown Oct 1 '18 at 17:32
• Instead of splitting the fractions the way you did, try $$\frac{\sin(x)(\cos(h) - 1)}{h} + \frac{\sin(h)\cos(x)}{h}$$ – Bungo Oct 1 '18 at 17:36

You broke up the initial limit incorrectly. The correct proof would be like this. (Notice how the limit is split up.) $$\sin’x = \lim_{h\to 0}\big(\frac{\sin(x+h)-sin(x)}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x\cos h- \sin x+\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x(\cos h-1)+\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\big(\frac{\sin x(\cos h-1)}{h}+\frac{\cos x\sin h}{h}\big)$$ $$=\lim_{h\to 0}\frac{\sin x(\cos h-1)}{h}+\lim_{h\to 0}\frac{\cos x\sin h}{h}$$ $$=\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}$$ $$=\sin x\cdot 0 + \cos x\cdot 1 = \cos x$$ $$\implies \boxed{\sin’x = \cos x}$$
In the second to last line, both limits don't exist. Akin to $\infty-\infty \neq 0$
$$\frac{\sin x\cos h+\cos x\sin h}{h}$$ in $$\frac{\sin x}h.$$