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!
Best Answer
" 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."
BINGO!
A limit existing, $\lim_{x\rightarrow a}f (x)=c $, means for every $\epsilon >0$ there is a $\delta $ so that whenever $|x-a|<\delta $ it also follows $|f (a)-c|<\epsilon $. That is impossible if $f $ is unbounded in all intervals around $0$.
Let $c $ be any real number. Clearly we can find $x $ close to $0$ so that $\frac 1{x^2} >c $.
To put it in terms of limits if $c $ is any arbitrary real number then for any $\epsilon >0$ we CAN'T find a $\delta $ so that $|x-0|=|x|<\delta $ would mean $|f (x)-c|=|\frac 1 {x^2} - c | < \epsilon $.
We can't do this as for any because for any $\epsilon $ and $c $ and any $\delta >0$ we can find $0 <x < \min (\sqrt {\max (c+\epsilon,\epsilon)},\delta) $ and $\frac 1 {x^2} > c+\epsilon $ so $|\frac 1 {x^2}-c|>\epsilon $ so there is no $\delta$ where the condition must be true.
So the limit can not be $c $. So the limit can't be any real number.