Limits of Transcendental Functions Determine the limit:
$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}$$
I know that the answer is $0$, I am just unsure of how to solve this. Thanks.
 A: You have different ways for doing it.
The first one is to write $$\cos(2h)=1-2\sin^2(h)\implies 1-\cos(2h)=2\sin^2(h)$$ $$\frac{1-\cos(2h)} h=\frac{2\sin^2(h)}h=2h\frac{\sin^2(h)}{h^2}=2h \left(\frac{\sin(h)}{h} \right)^2$$
The second one would be Taylor expansion
$$\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right)$$ Make $t=2h$ to get $$\cos(2h)=1-2 h^2+\frac{2 h^4}{3}+O\left(h^6\right)$$
$$\frac{1-\cos(2h)} h=2 h-\frac{2 h^3}{3}+O\left(h^5\right)$$ which shows the limit and how it is approached.
A: The third method is to multiply and divide by conjugate of numerator:
$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}=\lim_{h\to 0}\frac{1-\cos(2h)}{h} \cdot \frac{1+\cos(2h)}{1+\cos(2h)}=\lim_{h\to 0}\frac{\sin^2(2h)}{h(1+\cos(2h))}=$$
$$\lim_{h\to 0}\frac{\sin^2(2h)}{2h}=\lim_{h\to 0}\frac{\sin(2h)}{2h}\cdot \lim_{h\to 0} \sin(2h)=1\cdot0=0.$$
The fourth method is to use L'Hospital's rule:
$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}=\lim_{h\to 0}\frac{2\sin(2h)}{1}=0.$$
A: Let's go completely NUTS on this one. Let $2h=x$ so that the limit becomes$$2\lim_{x\to 0}\frac{1-\cos(x)}{x}$$ Now let $x=\pi/2-t$ so that the limit becomes$$2\lim_{t\to \pi/2}\frac{1-\cos(\pi/2-t)}{\pi/2-t}=2\lim_{t\to \pi/2}\frac{sint-1}{t-\pi/2}$$
Now apply the limit definition of the derivative for $sine$ at $t=\pi/2$ to arrive at $2\cos\pi/2=0$
