Why does $\frac{\cos\Delta x - 1 }{\Delta x} \to 0$ I'm watching Lecture 3 in MIT single variable calculus. 
https://www.youtube.com/watch?v=kCPVBl953eY&list=PL590CCC2BC5AF3BC1&index=3
And at one point the instructor does the following:

I was under the impression that when evaluating limits we need to avoid having $0/0$ in the denominator. However, in the notes here, it says that 
$\frac{\cos\Delta x - 1 }{\Delta x} \to 0$
How does this work?

hint
$$1-\cos (d)=2\sin^2(\frac {d}{2}) $$
$$|\cos (d)-1|\le \frac {d^2}{2} $$
if we know that
$$|\sin (A)|\le | A |$$
hence
$$\Bigl |\frac {\cos (\Delta x)-1}{\Delta x}\Bigr |\le \frac {|\Delta x|}{2} $$
