If $a_n \to 0^-$ . Why does $\frac{1}{a_n} \to -\infty$? If $a_n \to 0^-$ . Why does $\frac{1}{a_n} \to -\infty$ ?
I know that $\frac{1}{0} = \infty$ , of course, but why does a sequence tending to $0$ from the negative mean that the inverse tends to negative infintiy? why Negative?
 A: Note that
$$a_n\to0^-\implies a_n<0$$
Hence,
$$\frac{1}{a_n}<0$$
Therefore,
$$\frac{1}{a_n}\to-\infty$$
A: Have a look at https://www.desmos.com/calculator/cne7wuioah, particularly between $x=-1$ and $x=0$. You'll see the curve sloping down towards $-\infty$, just as it tends to $+\infty$ when $x$ tends to $0$ from the positive direction 
A: $\frac{1}{0} = \infty$ 
Actually, most mathematicians would not say that.  They would just say that is undefined.  
Read the definition of $\to -\infty$ carefully. It does not actually involve infinity at all.  It says that the sequence will be less that any negative value if you go beyond a certain point.  So, it will be less than negative million if you are beyond a suitable point, it will be less than negative billion beyond another suitable point, less than negative googolplex, etc.  Saying "tends to negative infinity" is just a suggestive name.  
A simple sequence that $\to -\infty$ is $-1, -2, -3, -4, -5, . . .$  It clearly satisfies the definition.  It will be less than $-N$ after the $N$th term.  No actual infinity (in any of its many senses) is involved.  
A: 
Lets try $x≈0.000000001$ and $x≈-0.000000001$.

I think, you understood the basic idea.
A: As $a_n$ approaches $0$, with $a_n \ne 0$, we get that $|a_n|$ approaches $0$ from above, so $\left|{\large{\frac{1}{a_n}}}\right|$ approaches infinity.

Hence if $a_n$ approaches $0$ from below, $\left|{\large{\frac{1}{a_n}}}\right|$ approaches infinity, but then, since each of the terms ${\large{\frac{1}{a_n}}}$ is negative, it follows that ${\large{\frac{1}{a_n}}}$ approaches minus infinity.

But note: Your assertion that ${\large{\frac{1}{0}}}=\infty$ is not correct.

A fixed version of that idea would be: ${\displaystyle{\lim_{x\to 0^+}}} {\large{\frac{1}{x}}}=\infty$.
