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?
 A: [This answer is more about how to read when using online sources for self-directed study than directly answering your question.]
You have had the solution on hand. The excerpt you put in the question is from the lecture note of Lecture 3. The professor explicitly mentions in the first page of the note that



You should refer back to the "last lecture" (Lecture 2)— it is done on page 10:





[Added later:]Remark. Note that this lecture note only gives a "plausible" argument why the results are true. You would learn in real analysis what a rigorous proof should look like and that would be another story. 
A: HINT: write your Quotient in the form $$\frac{(\cos(\Delta x)-1)(\cos(\Delta x)+1)}{\Delta x(\cos(\Delta x)+1)}$$
A: 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
$$\Bigl |\frac {\cos (\Delta x)-1}{\Delta x}\Bigr |\le \frac {|\Delta x|}{2} $$
A: 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
$$\lim_{\Delta x\rightarrow 0}\frac{\sin\Delta x}{\Delta x}\cdot \frac{\sin \Delta x}{\cos\Delta x+1}=1\cdot\frac{0}{2}=0.$$
A: Since you know that 
$$\lim_{x \rightarrow 0}\frac{\cos x -1}{x}=\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} $$
$$\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} =\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}$$
$$\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}$$
$$ \lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}=0$$
A: Suppose you know the limit exists.  Then by symmetry,
$$L=\lim_{x\to0}\frac{\cos(x)-1}x=\lim_{x\to0}\frac{\cos(-x)-1}{-x}$$
The second limit simplifies into
$$-\lim_{x\to0}\frac{\cos(x)-1}x$$
Finally, add these together to get
$$L+L=0\implies L=0$$
A: Using the fact that $\lim\limits_{\Delta x\to0}\frac{\sin(\Delta x)}{\Delta x}=1$ (as shown in this answer)
$$
\begin{align}
\lim_{\Delta x\to0}\frac{\cos(\Delta x)-1}{\Delta x}
&=\lim_{\Delta x\to0}\frac{\cos(\Delta x)-1}{\sin(\Delta x)}\frac{\sin(\Delta x)}{\Delta x}\\
&=\lim_{\Delta x\to0}\frac{\cos(\Delta x)-1}{\sin(\Delta x)}\lim_{\Delta x\to0}\frac{\sin(\Delta x)}{\Delta x}\\
&=-\lim_{\Delta x\to0}\frac{\sin(\Delta x)}{\cos(\Delta x)+1}\lim_{\Delta x\to0}\frac{\sin(\Delta x)}{\Delta x}\\
&=-\frac02\cdot1
\end{align}
$$
