M2C2A: Episode 16, Can we give a numerical value to 1/0?

Remember when some episodes ago we used P.I.G. (Patterns In Graphs)? We'll do it again.
What happens when you graph 1/x? You get a graph that looks like 2 rotated "L". You can see the closer x gets to 0, the closer y goes to ∞... but that if you look at the positive region!
For positive values of x, 1/x goes approaching ∞. For example:
1/1=1
1/0.1=10
1/0.01=100
...
But for negative values of x, 1/x goes approaching -∞. For example:
1/-1=-1
1/-0.1=-10
1/-0.01=-100
...
So which is it? ∞, or -∞?
It could be both.
Because 1/10^x=10^-x, then, 1/10^-∞ should be equal to 10^-∞ (And 10^-∞ isn't ∞. That would be like saying 10⁶ (1000000, a million) is equal to 6.), right? There are only 2 problems with that:
1) This could work for all bases, 1/10^x=10^-x is only a form of the more universal equation, 1/n^x=n^-x.
2) x^-∞ can't be 0... Or can it?
x¹=x, x⁰=1, but x^-1=1/x, and x^-2=1/x², so does x^-∞=1/x^∞ (which should definitely be larger than infinity if x>1)? Yes!
But what if you divide 1/<A number definitively bigger than infinity? It should give you a number with ∞ 0's...But is this 0? Think about this in this way, if this number had any digits after the 0's, it would have an end, and it would be infinite... So this is 0 after all! (hope I'm not breaking any kind of calculus law or something) But because x^∞ will only be infinite if x>1, the solution set for 1/0 would be 1/x^∞ only when x>1...
But our rule of 1/n^x=n^-x also applies to when x is positive...
So n^∞=1/n^-∞, But is n^-∞ also 0?
As you graph n^x, when x is negative, the values tend to be 0 as x goes down. So the only thing logical is that:
n^-∞=0!
So the solution set for 1/0 would be 1/x^∞ only when x>1 and x<-1.
-The Roaring Thunder