Epsilon-delta proof that $ \lim_{x\to 0} {1\over x^2}$ does not exist I'd like to see an epsilon delta proof that the $\lim: \lim_{x\to 0} {1\over x^2}$ does not exist and an explanation of the exact reason it does not exist, because I am not so sure I believe that the limit does not, in fact, exist, so I need to be proved wrong. 
What is the relationship between a limit existing, and the function in question having a least upper bound?  Because it seems to me that the only explanation I can find as to why the limit does not exist is that the function is unbounded.
I'm not sure why this is relevant because it seems to me that when $x$ approaches $0$ then ${1\over x^2}$ gets infinitely close to the y-axis which suggests to me that there does exist, in fact, an epsilon infinitely close to zero such that if $|x - a| < \delta$ then $|f(x)-L| < \epsilon$ where $\delta$ is infinitely close to zero and $\epsilon$ is infinitely close to zero. 
Obviously, my understanding of calculus hinges on this question, so I really need to be convinced with a bulletproof explanation, otherwise I'll continue to doubt the truth (I don't believe anything unless I fully understand it myself, for better or worse, I ignore other's authority and rely only on proof and logical understanding -- I'm sorry if this attitude offends anyone)!  Thanks in advance!

An epsilon-delta proof is used to show that the limit exists and is $L$, not usually to show that no limit exists.  We can see where it fails.  Suppose we claim that $\lim_{x \to 0}\frac 1{x^2}=L$  If somebody gives us an $\epsilon \gt 0$ we have to find a $\delta \gt 0$ such that $|x| \lt \delta \implies |f(x)-L|=|\frac 1{x^2}-L| \lt \epsilon$.  The problem is that if $x$ is very small, $\frac 1{x^2}$ is very large and can certainly be larger than $L+\epsilon$.  
You are confusing the fact that the graph of $y=\frac 1{x^2}$ gets close to the $y
$ axis, which really means $\lim_{x \to 0} x=0,$ with the value of $\frac 1{x^2}$ getting close to a value, which it does not.
