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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!

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    $\begingroup$ Let $f(x):=1/x^2$, then the limit exists for $f$ defined in $\bar{\Bbb R}:=\Bbb R\cup\{-\infty,\infty\}$, but it doesnt exists for $f$ defined in $\Bbb R$ because $\infty\notin\Bbb R$, and $\lim_{x\to 0}f(x)=\infty$. $\endgroup$ – Masacroso Mar 16 '17 at 21:15
  • $\begingroup$ "Then 1/x^2 gets infinitely close to the x axis". All functions get infinitely close to the x-axis as x gets infinitely close to 0. x getting close to 0 is synonymous with f (x) getting infinitely close to the y-axis (which is just the line x=0). You need that f (x) gets infinitely close to some y=L. It should be obvious the 1/x^2 "blows up" as x gets close to 0 so f (x) doesn't get close to any L (L is a value that f (x) gets close to. It's a number. Not a line and not the y-axis.). It gets bigger then all possible L. If L=1000, f (1/1000)-L= 999000 > epsilon. For example. $\endgroup$ – fleablood Mar 17 '17 at 6:01
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" 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.

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  • $\begingroup$ Interesting. I guess I just find it surprising we restrict ourselves in the real numbers by not admitting some notion of infinity as a real number (why not work in the extended real numbers by default?) but this must definitely be the reason the limit does not exist. $\endgroup$ – Robert Wilson III Mar 17 '17 at 10:27
  • $\begingroup$ I was thinking "why wouldn't L = f(x)?" but I guess that would require the extended real numbers to work. $\endgroup$ – Robert Wilson III Mar 17 '17 at 10:55
  • $\begingroup$ " I guess I just find it surprising we restrict ourselves in the real numbers by not admitting some notion of infinity as a real number" You find it surprising? I find it bloody obvious. Infinity isn't a consistent value. It doesn't obey basic arithmetic. No cauchy sequence converges to it. And there are no open neighborhoods around it, i.e. you can not get within epsilon of infinity. $\endgroup$ – fleablood Mar 17 '17 at 15:51
  • $\begingroup$ " "why wouldn't L = f(x)?" f(x) is not a single value. It is not a number. f(x) is the function itself. That concept doesn't make any sense. $\endgroup$ – fleablood Mar 17 '17 at 15:52
  • $\begingroup$ After thinking about it some more, I think you're right. It doesn't really mean anything. However, I have a natural propensity to want to say something more like L = the largest possible value of f(x) near 0. Just like I want to be able to say that ϵ = the smallest possible real number greater than 0. I guess this kind of thinking just misses the mark because people will respond that there is no such thing as these numbers. $\endgroup$ – Robert Wilson III Mar 18 '17 at 6:00
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I'll try to sketch the idea of the proof without giving all the details, since that part of the exercise is your job.

In order for the limit to exist there must some particular number $L$ for the limit. Then the function's value is will be close to $L$ whenever $x$ is close enough to $0$.

Since $1/x^2$ is very large when $x$ is near $0$, the limit $L$ will have to be a large number if it is to exist. Could the limit be $1000$? I don't think so. You should be able to show that if you look at values of $x$ close to $0$ then $1/x^2$ will be greater than $1001$, which tells you there's no $\delta$ that works with $\epsilon = 1$.

Now write the formal proof that no number $L$ can be the limit. So there is no limit.

PS Don't say that the limit is infinity. Infinity is not a number. There is a sense in which it's correct to say the limit is infinity, but this exercise does not ask about that and you can get into trouble if you try.

PPS Phrases like "infinitely close" sometimes provide useful intuition, and they were (in a sense) the best that mathematicians managed when calculus was being invented, but you should not use them now in formal arguments. The whole $\epsilon - \delta$ thing is designed to express the idea precisely. I think you can understand many of the ideas of calculus without it, but you can't prove theorems that way.

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  • $\begingroup$ By "infinitely close to 0" I really just mean any $\epsilon > 0$ in the real numbers, but in order for the definition of the limit to make any sense, the only important values of $\epsilon$ are the ones that are, idk, infinitely close to zero, because otherwise we could prove nearly anything was a limit if we weren't forced to check the smallest epsilon possible. $\endgroup$ – Robert Wilson III Mar 17 '17 at 10:50
  • $\begingroup$ There really is no such thing as "infinitely close to $0$" (unless you start into what's called nonstandard analysis), The whole point of the $\epsilon-\delta$ definition is to replace one statement about "infinitely close" to infinitely many statements, one for each actual genuinely positive $\epsilon$. It's OK to think the former, but you should only write the latter. $\endgroup$ – Ethan Bolker Mar 17 '17 at 13:12
  • $\begingroup$ Replace "gets infinitely close" with "as close as you like" or "arbitrarily close" and you have the general informal idea. A limit exists if when x gets arbitrarily close to 0, then f(x) gets artibtrarily close to a value. I think you get this. And I think it should be obvious that f(x) = 1/x^2, this most certainly does not happen. f(x) "blows up" instead and does not get close to any number. From you comments I see you were assuming that "f(x) approaches infinity" was an acceptable answer. It isn't. You can't get "close" to infinity. $\endgroup$ – fleablood Mar 17 '17 at 16:00
  • $\begingroup$ As for this idea of infinity as a limit, I've realized the difference between the real numbers and the extended real numbers. And according to Baby Rudin 1.23 and 4.33 in the extended reals I could set this up, since the upper bound for any subset of the reals would be +∞, every non-empty subset would have a lub. One that didn't have a lub in the reals, like the original function in the title of this question, would have lub = +∞ in the ext. reals, thus I could have f(x) near 0 be actual +∞. Then by 4.33 I could show that for any neighborhood around +∞ there exists one around x that works. $\endgroup$ – Robert Wilson III Mar 18 '17 at 23:52
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Do some examples. Let $L=57,345$. Let $\epsilon=.0000003$. If $|x| < .000,0001$ then $f (x) > 100000000000000$ so $|f (x)-L|> 99999999942655 > .0000003$.

It doesn't matter what delta we choose. If we choose a large delta we can still pick a very small $x $ to get a large $f (x)-L $. And if we pick a small $\delta $ it only makes $|f (x)-L|$ larger.

And no matter how large we make $L $ we can always find a small enough $x $ to make $f (x)-L$ very large rather than small. Example. Let $L=10^{(10^{100})} $ and $0 <\epsilon <1$. We simply have to pick $x<10^{10^{100}} $ and $f (x)-L=10^{2*10^{100}}-10^{10^{100}}=10^{10^{100}}(10^{10^{100}} -1)$ which last time a checked was a teeny bit more than $1$ which is larger than $\epsilon $

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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.

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  • $\begingroup$ Would you not call an "epsion-N" or a "can always be bigger than epsilon" proof of the non existence of limit or non-convergence f a sequence a "delta-epsilon" proof. i.e. We can and frequently do a proof that for any epsilon and any c then for all delta there will always be an x so that |x-a|<d and |f(x) - c|> e. Woulld that not be a delta-epsilon proof? Perhaps it wouldn't because the delta is not relevant and frequently omitted. $\endgroup$ – fleablood Mar 17 '17 at 16:09
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I think you have the idea backwards. You are wanting to know if the $y$ values of the function converge to something or not as $x$ approaches $0$. That epsilon part, $|f(x)-L|$ is looking at the $y$ values, not the $x$-values. It is clear that as $x \to 0$, $f(x)=1/x^2$ becomes increasingly big. For instance if $x=1/10$, $f(\frac{1}{10})=100$. If $x=1/100$, $f(\frac{1}{100})=10000$, etc. Therefore, there is no fixed $L\in \mathbb{R}$ for which it could converge. This limit is infinity.

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