I'm looking at a proof in a textbook that shows that $ \lim_{x \to 0}\frac{1}{x} $ does not exist. This is a paraphrase of the proof:

So, we want to show that there exists an $\epsilon>0$, such that for each $\delta >0$, there is some $x \in \mathbb{R}-{0}$, such that $|x-0| < \delta$, but $|\frac{1}{x}-L|\geq \epsilon.$ There are three cases, depending upon whether $L > 0$, $L = 0$, or $L < 0$. Suppose $L>0$. Then we have $\frac{1}{x} - L\geq \epsilon$, which is the same as $\frac{1}{x} \geq L + \epsilon$, and so we need $x \leq \frac{1}{L+\epsilon}$. Since $L >0$, the last step works with any choice of $\epsilon$. We also need to have $|x-0| < \delta$, and so we will choose $x = min\left\{\frac{\delta}{2},\frac{1}{L+\epsilon}\right\}$. We will not give the full details of the cases where $L < 0$ and where $L = 0$, except to note that when $L < 0$ we need to choose some $\epsilon > 0$ so that $L+ \epsilon < 0$, for example we could choose $\epsilon = \frac{|L|}{2}$.

For the life of me I can't understand why for the case of $L<0$, we must have $L+\epsilon <0$. I've tried playing around with the inequality and plugging in real numbers, and nothing is clicking. I'm sure it's something simple, but I just can't see it. Please know that I have looked at other proofs for this specific limit (or lack thereof) on here, and it still doesn't make sense.

This is the proof they did for $L>0$: Suppose that $\lim_{x\to 0} \frac{1}{x} = L$ for some $L \in \mathbb{R}.$ Let $\epsilon = \frac{|L|}{2}$ if $L \neq 0$, and let $\epsilon = 1$ if $L = 0.$ We consider the case when L > 0; the other cases are similar,and the details are left to the reader. Let $\delta > 0.$ Because $L > 0$, then $L+\epsilon > 0$. Let $x = min\left\{\frac{\delta}{2},\frac{1}{L+\epsilon}\right\}.$ Then $x \in (0,\infty)$ and $|x-0| \leq \frac{\delta}{2} < \delta.$ On the other hand, because $x\leq \frac{1}{L+\epsilon}$, it follows that $L + \epsilon \leq \frac{1}{x}$, and hence $\frac{1}{x} - L \geq \epsilon.$

Okay, so this makes sense to me. But could anyone help me with the proof for the case of $L<0$. I know there are probably better ways, but I want to understand within the context of the book. This is not homework, but self-study.

Thank you!

  • $\begingroup$ It's not that all $\epsilon$ are such that we have $L +\epsilon <0$. It's that of all the $\epsilon$ we can choose (and we can choose any $\epsilon \in (0, \infty)$. There are some we could choose where $L + \epsilon < 0$. For example why don't we just choose $\epsilon =\frac {|L|}2$. That $>0$ and less than $|L$ so $L +\epsilon -L + \frac {|L|}2 = \frac {L}{2} < 2$, isn't it. Or we could choose $\epsilon = 10^{-100}\cdot |L|$. Then $L +\epsilon =L +10^{-100}|L| = (1-10^{100})L < 0$. $\endgroup$
    – fleablood
    Mar 30, 2021 at 1:52
  • $\begingroup$ Thank you, fleabood. The textbook's insistence on the constraint $L + \epsilon <0$ really threw me off. These $\epsilon-\delta$ proofs get confusing because the $\epsilon$ sometimes seem to appear out of nowhere. $\endgroup$
    – dorkichar
    Mar 30, 2021 at 21:29

3 Answers 3


Since you are interested in a $\epsilon, \delta$ proof, when $L < 0$, this is how I would approach it.

Suppose that $L < 0$. Then I have to find some fixed value of $\epsilon > 0$, so that no matter what value of $\delta > 0$ is chosen, I will always be able to find at least one value of $x$ such that both of the following are true:

$0 < |x - 0| < \delta$ and $|L - (1/x)| > \epsilon$.

I will arbitrarily choose $\epsilon = 2|L| = -2L,$ since $L < 0.$

Since I want the normal constraint against $\epsilon$ violated:
I want $|L - (1/x)| > -2L$.

Here, to simplify matters, I will arbitrarily decide that I will find a value of $x$ that is positive and such that
$0 < |x - 0| < \delta \implies 0 < x < \delta.$

Also, since $x$ is positive, $|L - (1/x)| = (1/x) - L.$

Therefore, I have to prove that no matter how small (positive) $\delta$ is taken, I can find $x$ such that

$0 < x < \delta$ and $(1/x) - L > -2L \implies 1/x > -L \implies x < -1/L$.

Take $x = \min[(\delta/2), (-1/2L)]$, and you have found a satisfactory $x$. Since you will always be able to find such a satisfactory $x$, regardless of how small positive $\delta$ is taken, the normal constraint on the specific value of $\epsilon = -2L$ will be violated.

This refutes the assumption that as $x \to 0, (1/x) \to L$.

This is probably the opportune moment to mention that I pulled a fast one. My analysis started with a constraint to be violated, and then said that if it is violated, then
$x = \min[(\delta/2), (-1/2L)]$ appears to work.

To be rigorous, you need to manually check that the chosen value of $x$ is satisfactory. This step is usually skipped in continuity proofs, but is still a good idea.

Take $x = \min[(\delta/2), (-1/2L)].$

Then, $0 < x \leq (\delta/2) \implies 0 < |x - 0| < \delta.$

Also $0 < x \implies |L - (1/x)| = (1/x) - L.$

$0 < x \leq (-1/2L) \implies 0 < x < -(1/L) \implies (1/x) > -L \implies 1/x - L > -2L$, as required.

An explanation of my overall thinking is called for. In a general demonstration of this type, the logic usually goes:

Constraints $\implies$
Step 1 $\implies$
Step 2 $\implies$
Step 3 $\implies$
Step 4 $\implies$
Step 5 $\implies$

This means that if the Constraint is satisfied, the Specification is implied. However, you normally want the reverse implication that if the Specification is followed, the Constraint will be satisfied.

This is normally handled in one of three ways.

  • blithely assume that all implications were two way implications (i.e. $\iff$ instead of $\implies$). This is in fact almost always true.

  • Actually go to the trouble of checking each step to manually verify that the two way (i.e. $\iff$) implication held from the previous step to this step.

  • Take the very unusual approach that I took of treating the specification as a candidate specification and manually verifying that it works.

Responding to the Paramanand Singh's comments following this answer.
If I understand your strategy for showing/disproving continuity, it may be represented by:

Constraint : $~\impliedby$
Step 1 : $~\impliedby$
Step 2 : $~\impliedby$
Step 3 : $~\impliedby$
Step 4 : $~\impliedby$
Step 5 : $~\impliedby$

I agree that your strategy for showing/disproving continuity is valid.
However, I disagree that it is the only valid approach or that it is the easiest valid approach.

My approach is as follows:

Constraint : $~\implies$ or suggests
Step 1 : $~\implies~$ or suggests
Step 2 : $~\implies$ or suggests
Step 3 : $~\implies$ or suggests
Step 4 : $~\implies$ or suggests
Step 5 : $~\implies$ or suggests
Candidate Specification.

Then, I manually verify that Candidate Specification $~\implies~$ Constraint.
This approach is also valid, because of the trailing manual verification.

The mathSE query that you linked to involved showing that
$~\lim_{x \to 3} x^2 = 9.$
I will illustrate my alternative approach with this particular problem.

Constraint: $~0 < |x -3 | < \delta \implies |x^2 - 9| < \epsilon$.

Derived (i.e. suggested or implied) steps:

$- \delta < x -3 < \delta.$

$3 - \delta < x < 3 + \delta. $

$9 - 6\delta + \delta^2 < x^2 < 9 + 6\delta + \delta^2.$

$ - 6\delta + \delta^2 < (x^2 - 9) < 6\delta + \delta^2.$

To simplify the above suggested (i.e. derived) step, I will introduce the artificial contrivance that $\delta \leq 1.$

This will allow me to assume that $\delta^2 < \delta.$

This leads to the derived step:
$-6 \delta < -6\delta + \delta^2 < (x^2 - 9) < 6\delta + \delta^2 < 7\delta.$

What I want is that
$-\epsilon < (x^2 - 9) < \epsilon$.

The above analysis suggests that this is achieved by having
$\delta \leq 1$ and $\delta < (\epsilon/7)$.

This is achieved by the Candidate Specification of $\delta = \min(1, \epsilon/10)$.

It now remains to manually verify that the Candidate Specification does in fact imply that $0 < |x-3| < \delta \implies |x^2 - 9| < \epsilon$.

This is equivalent to verifying that when $x \neq 3$ and
$- \delta < (x-3) < \delta$, then $-\epsilon < (x^2 - 9) < \epsilon.$

Test the Candidate Specification:
$\delta = \min(1, \epsilon/10)$.

This specification, along with the presumption that $0 < |x-3| < \delta$ implies the following:

$\delta \leq 1~~$ and $~~\delta \leq (\epsilon/10).$
$x \neq 3~~$ and $~~ [-\delta < (x-3) < \delta] ~~\implies~~ [3 - \delta < x < 3 + \delta].$

$9 - 6\delta + \delta^2 < x^2 < 9 + 6\delta + \delta^2 \implies $
$- 6\delta + \delta^2 < (x^2 - 9) < 6\delta + \delta^2 \implies$
[because $\delta \leq 1$]
$-6 \delta < (x^2 - 9) < 7\delta \implies$
[because $\delta \leq (\epsilon/10)$]
$-\epsilon < (x^2 - 9) < \epsilon,~~$ as required.

Notice that very little thought was involved in the manual verification of the Candidate Specification. This is because the path from the Specification to the Constraint very closely followed the original derived analysis. Further, because of this trailing manual verification, there was no need to be concerned whether the reverse implication (i.e. $\impliedby$) held in each step in the derived chain of steps.

This means that I was free to ignore any consideration of the validity of going from one derived step to the next derived step. Establishing the validity was deferred until a Candidate Specification was computed.

  • $\begingroup$ Why are you trying to think limit definition as a two way implication? This is one of the biggest problems students face by assuming two way implication. $\endgroup$
    – Paramanand Singh
    Mar 30, 2021 at 1:43
  • $\begingroup$ @ParamanandSingh Yes, it is a common mistake, which was the whole point of my rant, at the tail end of my answer. But the fact is, normal algebraic manipulations are generally two way implications; the typical non-2-way implication is something like $x = 2 \implies x^2 = 4$. Further, the OP needs to know why this common mistake is so often committed; in fact, the OP needs to know that there is a potential mistake there. My answer indicated that. $\endgroup$ Mar 30, 2021 at 1:46
  • $\begingroup$ @ParamanandSingh Actually, I'm not sure that my last comment was on point to your comment. You do normally have a Constraint, and you have to derive a specification that satisfies the constraint. The normal derivation strategy does involve something like if the Constraint is satisfied, then Step 1 is implied, which implies Step2, which implies ... $\endgroup$ Mar 30, 2021 at 1:49
  • $\begingroup$ @ParamanandSingh Do you disagree? $\endgroup$ Mar 30, 2021 at 1:50
  • $\begingroup$ I view the overall $\epsilon, \delta$ proof to be written as "$q$ holds if $p$ holds" instead of "if $p$ then $q$". Both these are mathematically same but the first approach is what allows us to write the chain of logical implications in reverse manner. Starting with a target inequality $|f(x) - L|<\epsilon $ (or $|f(x) - L|\geq \epsilon$ for disproving limit) we can form a chain of logical implications in reverse order to arrive at our $\delta $ (or an $x$ for every $\delta$ in case of disproof). Noting that each step is a one way implication (in reverse) gives us a lot of freedom. $\endgroup$
    – Paramanand Singh
    Mar 30, 2021 at 5:25

If $L < 0$ and $\epsilon = \frac{|L|}{2}$ then \begin{align*} L + \epsilon &= L + \frac{|L|}{2} \\ &= L + \frac{-L}{2} \qquad\text{(because $L<0$)} \\ &= \frac{2L-L}{2} \\ &= \frac{L}{2} \\ &= \frac{1}{2} L \\ &< 0 \qquad\text{(because a positive times a negative is negative)} \end{align*} If that's not intuitive enough, here's a more geometrical viewpoint.

$L$ is negative, i.e. it lies to the left of $0$. The distance between $L$ and $0$ equals $|L|$. If you add only half that distance to $L$ then what you get still lies to the left of zero. That is, $L + \frac{|L|}{2} < 0$.

  • $\begingroup$ Thank you for sharing your answer, Lee. I understand why $L + \epsilon <0$ if $\epsilon = \frac{|L|}{2}$, but why do they insist that you must pick an $\epsilon$ such that $L + \epsilon <0$? How is that related to the inequality? Sorry if I'm not explaining my confusion well. $\endgroup$
    – dorkichar
    Mar 30, 2021 at 0:46
  • $\begingroup$ Your use of "must" is indeed not well chosen. There is no "must". One chooses $\epsilon = \frac{|L|}{2}$ because it works. What it means to "work" is that using that value of $\epsilon$ one can prove what one is supposed to prove, namely: for each $\delta > 0$ there exists $x \in \mathbb R - \{0\}$ such that $|x-0| < \delta$ but $|\frac{1}{x} - L| > \epsilon$. $\endgroup$
    – Lee Mosher
    Mar 30, 2021 at 1:19
  • 2
    $\begingroup$ @dorkichar: your textbook authors perhaps wants to say that if you want to have a proof for $L<0$ similar to the case $L>0$ then you must use $\epsilon$ such that $L+\epsilon<0$. $\endgroup$
    – Paramanand Singh
    Mar 30, 2021 at 1:48
  • 1
    $\begingroup$ It's not that you must choose and $\epsilon$ so that $L + \epsilon < 0$. It's that you could choose such. And if you do choose one the you willhave $f(x) < 0$. Which could make things easier. If you want to show that $|L - \frac 1x|> \epsilon$ and you can have $x$ go all the way to $0$ and past it's not lickly you can say anything. But if have it guarenteened that $x < 0$ then $L -\frac 1x$ is positive and manipuliably large. $\endgroup$
    – fleablood
    Mar 30, 2021 at 1:58
  • $\begingroup$ @fleablood you addressed exactly where my confusion was. Thank you! $\endgroup$
    – dorkichar
    Apr 4, 2021 at 20:58

The proof in your book seems to be unnecessarily complicated: as $x \to 0$, $|\frac{1}{x}|$ increases without bound, so $\frac{1}{x}$ cannot have a limit in $\Bbb{R}$.

You can extract an $\varepsilon$-$\delta$ proof from the above. Given any $L \in \Bbb{R}$, if $\epsilon = 1$ and $\delta = \frac{1}{|L| + 1}$, then $|\frac{1}{x} - L| > \epsilon$ for all $x$ with $-\delta < x < \delta$. This means that $\frac{1}{x} \not\to L$ as $x \to 0$


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