Prove that:

$$\lim_{x \to 0} \frac{1}{x}$$

Is non existent.

This is my attempt:

Assume The limit $= L$ Some real number $L$

By the definition of one sided limits we get from right and left respectively:

For $\delta > 0, \epsilon > 0$ $$x< \delta \implies \left| 1/x - L \right| < \epsilon$$

$$-x < \delta_2 \implies \left| 1/x - L \right| < \epsilon$$

Let $\delta' = \min(\delta_1, \delta_2)$

Let $\epsilon = 1$

$$x< \delta \implies \left| 1/x - L \right| < 1$$

$$-x < \delta_2 \implies \left| 1/x - L \right| < 1$$

$$0 < \delta_1 + \delta_2 \implies 2|1/x - L| < 2$$

what else should I do?

  • $\begingroup$ Your definition of the one-sided limit is wrong. "There exists" and "for all" are different. They should not be confused, and the order cannot be changed. $\lim_{x\to a} f(x)=L$ iff for all $\epsilon>0$, there exists $\delta>0$ s.t. for all $x\in \mathbb{R}$, $0<|x-a|<\delta\implies |f(x)-L|<\epsilon$. For left/right limit, you just have $x$ further restricted to the left/right of $a$. $\endgroup$ Dec 8, 2014 at 13:09

3 Answers 3


This is my first answer in Stackexchange. I hope that my answer can give you some help. It's said that the limit of $f(x)$ is L when x approaches to $0$ if the following statement is true:

For every $\delta>0$, there exists $x_0>0$ such that for $\left|x\right|<x_0$ implies that $\left| f(x)-L \right|<\delta$.

Here we assume this statement holds for $f(x)=\frac{1}{x}$. There exists L, that for every $\delta>0$, there exists $x_0$ such that for $\left|x\right|<x_0$ implies that $\left| \frac{1}{x}-L \right|<\delta$.

Given $\delta$ and $L$, apply them to the inequality:$$\left| \frac{1}{x}-L \right|<\delta$$

$$\frac{1}{x}-L<\delta$$ $$x>\frac{1}{L+\delta}$$

Just consider the case when $L>-\delta$. Here needs some attention, because if we want to get a paradox for a statement with every or any, we just give an example where the statement is not true, then the whole statement is not true.

So that if $0<x<\min(x_0,\frac{1}{L+\delta})$, $\left| \frac{1}{x}-L \right|>\delta$ which is contrast to the original statement because we find a $x$, whose absolute value is smaller than $x_0$ while not satisfied with the inequality.

Proof is finished.

  • $\begingroup$ How can you consider a specific $L$ case? $\endgroup$
    – Amad27
    Dec 8, 2014 at 14:50
  • $\begingroup$ @Amad27 I don't specify L, I just specify $\delta$, because my goal is to prove some(not any) $\delta$ that is not satisfied with that statement under the assumption of L's existence. So a specific $\delta$ is considered and then we can conclude that L does not exists. Here involves a series of logical reasoning. $\endgroup$
    – Vassily
    Dec 9, 2014 at 8:24

At the very least, this proof needs a lot more explanation. A mathematical proof is not just a series of symbols: it is a flow of ideas, and ideas are best conveyed using both symbols and SENTENCES.

For example, you say you assume that the limit is $L$. OK, now I understand what $L$ is, but what are $\delta, \delta_2$ and $x$? And what is $\epsilon$? Explain these things in sentences and make it clear what you are trying to prove, what follows from what, and, most importantly, why this means that the limit does not exist.

A hint:

The statement

$\lim_{x\to x_0} f(x)$ exitsts

is equivalent to saying:

There exists such a $L$ that for every $\epsilon > 0$, there exists such a $\delta > 0$ that from $|x-x_0|<\delta$, it follows that $|f(x)-L|<\epsilon$.

Therefore, you must prove that the limit does not exist, meaning that this statement is not true. What is the negation of this statement?

As I see you are a beginner, I will help a little more:

To prove that $f$ has no limit, you must prove that for every $L$, $L$ is not a limit, meaning that there, for every $\delta > 0$, there exists such an $x$ that $|x-x_0| < \delta$ and $|f(x) - L| > \epsilon$.

  • $\begingroup$ I didnt prove it, I need help proving it. $\endgroup$
    – Amad27
    Dec 8, 2014 at 12:57
  • 1
    $\begingroup$ @Amad27 I didn't know that. You know why I didn't know that? Because I have no idea what you were trying to do. Please, explain what you were trying to do using both sentences and symbols. All my comments still stand. $\endgroup$
    – 5xum
    Dec 8, 2014 at 12:58
  • $\begingroup$ Sorry, Ill update the question. $\endgroup$
    – Amad27
    Dec 8, 2014 at 12:58
  • $\begingroup$ @Amad27 Still not good. Did you pick an $\epsilon$? Are you saying that you take an arbitrary $\epsilon$? $\endgroup$
    – 5xum
    Dec 8, 2014 at 13:03
  • $\begingroup$ Am I supposed to pick an $\epsilon$ $\endgroup$
    – Amad27
    Dec 8, 2014 at 13:12

There is a limit $L$ at $0$ means that for values of $x$ sufficiently close to $0$, $f(x)$ remains close to $L$.

But in the case of the inverse function $1/x$, you will always find values of $x$ close to $0$ that have their inverse far from $L$.

Take $x=1/(L+D)$, you can make it tiny, and at the same time far from $L$ as $1/x-L=D$.


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