I need help understanding how this limit is proved? :
Show that $$\lim_{h\to 0} \frac{\cos (h)-1}{h}=0$$
Proof:
Using the half angle formula, $\cos h = 1-2 \sin^2(h/2)$
$$\lim_{h\to 0} \frac{\cos (h)-1}{h}\\=\lim_{h\to 0}( -\frac{2 \sin^2(h/2)}{h})\\=-\lim_{\theta \to 0}\frac{\sin \theta}{\theta} \sin \theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Let $\theta=h/2$)} \\ = -(1)(0)\\=0$$
I have no idea how this proof is done, so I apologize for the lack of my own thoughts in this question. I understand limits and know sin, cos, tan, but I am just very lost as what they did in each step. Can someone please explain all the steps of the proof as well as the half-angle formula. Thanks!
Best Answer
The simplest proof is this: $$\frac{\cos h-1}h=\frac{(\cos h-1)(\cos h+1)}{(\cos h+1)h}=\frac{\cos^2h-1}{(\cos h+1)h}=-\frac{\sin^2h}{(\cos h+1)h}=-\frac{\sin h}h\cdot\frac{\sin h}{\cos h+1}.$$ The first fraction tends to $1$, the second tends to $\dfrac 02=0$, hence the limit is $\color{red}0$.
For the proof you mention, at the third line, you should have $$=\lim_{h\to 0}\Bigl( -\frac{2 \sin^2(h/2)}{h}\Bigr)=\lim_{h\to 0}\Bigl( -\frac{\sin^2(h/2)}{h/2}\Bigr)=\dots$$