<s>[INST] 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?
 [/INST] Since you know that 
$$\lim_{x \rightarrow 0}\frac{\cos x -1}{x}=\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} $$
$$\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} =\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}$$
$$\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}$$
$$ \lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}=0$$
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