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?

By the mean value theorem,
$$\frac{\cos\Delta x-1}{\Delta x}=\frac{\cos\Delta x-\cos(0)}{\Delta x-0}=-\sin(c)$$
for some $c\in (0,\Delta x)$
Let $\Delta x \rightarrow0$ and see what happens.
An alternative way of computing the limit:
$$\frac{\cos\Delta x-1}{\Delta x}=\frac{\cos^2\Delta x-1}{\Delta x\cdot(\cos\Delta x+1)}=\frac{\sin^2\Delta x}{\Delta x\cdot(\cos\Delta x+1)}=\frac{\sin\Delta x}{\Delta x}\cdot \frac{\sin \Delta x}{\cos\Delta x+1}$$
Now, as $\Delta x\rightarrow0$, by continuity and the known fact that
$$\lim_{\Delta x\rightarrow 0}\frac{\sin\Delta x}{\Delta x}=1$$
we get
$$\lim_{\Delta x\rightarrow 0}\frac{\sin\Delta x}{\Delta x}\cdot \frac{\sin \Delta x}{\cos\Delta x+1}=1\cdot\frac{0}{2}=0.$$
