Proof for nonexistent of $\lim_{x\to 0}\frac{1}{x}$

Question

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?

Answer 1

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.

Answer 2

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.

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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 $|f(x) - L| > \epsilon$.

Answer 3

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 $1/x-L=D$.