Recently I took a test where I was given these two limits to evaluate:
$\lim_\limits{h \to 0}\frac{\sin(x+h)-\sin{(x)}}{h}$ and $\lim_\limits{h \to 0}\frac{\cos(x+h)-\cos{(x)}}{h}.$ I used sine and cosine addition formulas and found the value of each limit individually, eventually canceling out $\sin x\cdot \frac1h$ and $\cos x\cdot \frac1h$ because I learned that we can evaluate limits piece by piece. As a result, I got $\cos x $ and $-\sin x$ as my two answers. However, my teacher marked it wrong saying that we cannot cancel $\sin{x}\cdot\frac1h$ or $\cos{x}\cdot\frac1h$ because those limits do not exist. Can someone please explain why this doesn't work? I thought that we can cancel those limits out since we never look at $0,$ just around $0,$ when evaluating these two limits.
Sine: $$\frac{\sin(x+h)-\sin(x)}h=\frac{\sin(x)\cos(h)+\sin(h)\cos(x)}h-\frac{\sin(x)}h$$
$$=\sin(x)\frac1h+\cos(x)-\sin(x)\frac1h=\cos(x)$$
Cosine: $$\frac{\cos(x+h)-\cos(x)}h=\frac{\cos(x)\cos(h)-\sin(x)\sin(h)}h-\cos(x)\frac1h$$
$$=\cos(x)\frac1h-\sin(x)\cdot1-\cos(x)\frac1h=-\sin(x)$$
Note: I am allowed to assume $\lim_\limits{x\to 0} \frac{\sin(h)}h=1,\lim_\limits{x\to 0} \frac{\cos(h)-1}h=0.$
Best Answer
That's the standard way to find the derivatives for $\sin x$ and $\cos x$ by the definition.
Your way was not correct since as $h\to 0$ we have $\sin x/h \not \to 0$, to proceed properly for the first one we have
$$\sin(x+h)-\sin{(x)}=\sin x\cos h+\sin h\cos x-\sin x$$
and then
$$\lim_{h \to 0}\frac{\sin(x+h)-\sin{(x)}}{h}=\lim_{h \to 0}\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}$$
$$=\sin x \cdot\lim_{h \to 0}\frac{\cos h-1}{h}+\cos x\lim_{h \to 0}\frac{\sin h}{h}$$
and from here to conclude we need to use standard limits
For the other one we can proceed in a similar way.