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I would like to understand how to complete this proof, for I am unable to conclude. Also I wanted to know if what I wrote before is correct. I tried to follow an answer over here (this one: How to prove a limit exists using the $\epsilon$-$\delta$ definition of a limit and the answer I followed is by the user FREETRADER).

So I want to prove $\lim_{x\to 2} 3x = 7$ is false.

Following the linked example, let $\epsilon > 0$. We have to demonstrate that there does exists $\delta > 0$ such that IF $0 < |x-2| < \delta$ THEN $|f(x) - \ell| < \epsilon$.

In what way the size of $|x-2|$ afflicts the size of $|f(x) - 7|$? Since $f(x) = 3x$ then we have $|3x-7|$. Now we have to produce a way to be able to tell how much $f(x)$ gets close to $7$ (that is we choose an $\epsilon$), and from here we get how much $x$ has to be close to $2$ so that is true (hence we get $\delta$ which makes it true).

$$0 < |x-2| < \delta \iff -\delta < x-2 < \delta \iff 2-\delta < x < 2+\delta$$

So I write

$$f(2\pm \delta) = 6 \pm 3\delta$$

And then

$$|f(x)-7| < \epsilon \implies |6\pm 3\delta -7| < \epsilon$$

This means $|-1 \pm 3\delta | < \epsilon$.

Now I got stuck because the above expression also reads

$$\frac{1-\epsilon}{3} < \pm \delta < \frac{1+\epsilon}{3}$$

I tried to separate into the two signs of $\delta$ but I don't understand how to conclude that this gives some kind of contradiction, or absurd, or well how this tells the limit is false...

Thank you!

Gerr
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  • May be try to show, that negation is true. – zkutch Oct 04 '24 at 10:15
  • $0 < |x-2| < \delta \iff -\delta < x-2 < \delta$ This is wrong. – Bowei Tang Oct 04 '24 at 10:15
  • @BoweiTang You're right, it suffices $x - 2 < \delta$ since $x-2 \geq 0$. But the problem remains, in the end – Gerr Oct 04 '24 at 10:27
  • You need a sequence converging to 2 for which the function remains outside a certain interval of 7. – zkutch Oct 04 '24 at 10:30
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    You need to use negation of limit definition and that involves finding a suitable value of $\epsilon$ such that limit definition does not hold for nay value of $\delta$. A typical approach is to guess actual limit ($6$ here) and then take $\epsilon$ to be less than $|\text{actual limit} - \text{false limit} |$. – Paramanand Singh Oct 04 '24 at 10:31

3 Answers3

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Your mistake is in the first step, where you wrote "let $\epsilon > 0$." You seem to be trying to prove that for every $\epsilon > 0$ there does not exist a $\delta > 0$ such that .... But that's not what you have to prove (and in fact it's not true). What you have to prove is that there is at least one $\epsilon > 0$ such that there does not exist a $\delta > 0$ such that .... The easiest way to do this is to find an example of an $\epsilon$ for which there is no corresponding $\delta$. Your final solution should start by specifying a particular number $\epsilon$, and then you should show that, for that particular value of $\epsilon$, there is no value of $\delta$ that works.

Dan Velleman
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When you would like to to prove, that $\lim\limits_{x\to 2} 3x = 7$ is false it is same to prove, that its negation is true. Negation of brought fact is $$\exists \varepsilon>0, \forall \delta>0, \exists x\in U_{\delta}(2)\setminus\{2\},|f(x)-7|=|3x-7|\geqslant \varepsilon$$ if, for example, we consider $\delta_n=\frac{1}{n}$ and $x_n=2+\frac{1}{2n}$, then $|x_n-2|=\frac{1}{2n}<\frac{1}{n}$ and in same time $|3x_n-7|=\left | \frac{3}{2n} -1 \right| \geqslant \frac{1}{4}$.

zkutch
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  • Let me add that, as I feel it, there is no "other" way than to prove the truth of negation. Any other way may be well or badly disguised but still prove precisely negation - the essence of negation, somewhat artistically formulated, is that the variables under the universal quantifier are transformed into fixed constants, and fixed quantities, under existential quantifier, become variables with respect to which one must work. – zkutch Oct 04 '24 at 15:53
  • Thank you for this. Just a question: how did you get that the last inequality is $\geq \frac{1}{4}$? I miss that part... – Gerr Oct 05 '24 at 20:45
  • Added one equality more. – zkutch Oct 05 '24 at 20:54
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I think the easiest way to think of this is you need to show that as $x$ "gets close" to $2$ then $3x$ does not get close to $7$. Which is obvious as $3x$ "gets close" to $6$. So as $x$ gets close to $2$ we have $3x$ getting further away from $7$.

For example. We can have $|3x -7| > \frac 12$ if $6.5< 3x < 7.5$ or in other words if $2\frac 16 < x < 2\frac 12$. In this case $|x-2| > \frac 16$.

Why is this a contradiction? Because if we chose $\epsilon =\frac 12$ then we can not find a $\delta>0$ so that $|x-2|<\delta$ implies $|3x-7| < \epsilon = \frac 12$. Why not? Because for any $\delta > 0$ we can find an $x$ so that $|x-2| < \min(\frac 16, \delta) \le \delta$ but $|3x-7| > \epsilon =\frac 12$.

To make this not so specific and more formal...

For numbers very close to $2$, that is $|x-2| < \delta$ for some small $\delta$ we have $2-\delta < x < 2+\delta$ and $6-3\delta < 3x < 6+3\delta$ and $-1-3\delta < 3x-7 < -1 +3\delta$. And for a $\delta$ small enough so that $-1+3\delta < 0$ (i.e. $\delta < \frac 13$ we get $|3x-7| > 1-3\delta$.

How is this a contradiction? Well. Let $0 < \epsilon < 1$ and let $\delta_1$ be such that $\epsilon = 1-3\delta_1$ or $\delta_1 = \frac 13 -\frac \epsilon 3$. Then for any $\delta$ if we choose an $x$ so that $|x-2|< \min(\delta, \delta_1)<\delta $ we will have $|3x-7|> \epsilon$.

Thus it is not true that for any $\epsilon$ there exists $\delta$ so that for all $x$ so that $|x-2| < \delta$ we have $|3x-7| < \epsilon$.

(...because.... just to beat a dead horse.... for an $\epsilon: 0 < \epsilon < 1$ and any $\delta$ we have for any $x: |x-2| < \min(\delta, \frac 13-\frac \epsilon 3)$ that $|3x-7| > \epsilon$.)

fleablood
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