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Suppose I have a wrongly evaluated limit$-$ $$\lim_{x \to2} 3x+3=6$$Then we get $-$ $$3x+3-6<\varepsilon $$ $$3x-3<\varepsilon $$$$|x-1|<\frac\varepsilon {3}$$$$|x-2|<\frac\varepsilon 3 -1$$ $$|x-2|<\frac{\varepsilon-3} 3=\delta $$ What is wrong with this choice of $\delta$? (Also, please try to provide a diagram, which will help beginners like me!)


P.S I had posted another question, based on this question (which was closed recently). I wrote this post making the question explicit, so would expect a decent answer.Thanks!

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    $\begingroup$ the part where you go from |x-1| to |x-2| is false, on the RHS you have a negative quantity for a small epsilon. $\endgroup$ Commented Feb 24, 2021 at 11:22

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When you write $|x-2| < \varepsilon/3-1$ this means $$ 0\le|x-2| < \dfrac{\varepsilon-3}{3}\\ \varepsilon \ge 3 $$ For this inequality to hold true

But in the limit definition we require $ \forall \epsilon > 0$

EDIT

To prove that a sequence don't have a limit or in other terms it diverges we need to show that there doesn't exist any real number $l$ such that the limit definition holds which is the logical negation of the statement.

The above result is as general as the epsilon defition as it is just the negation of it.

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  • $\begingroup$ Your answer is good, but could you please add a diagram, for a geometric approach? $\endgroup$
    – Eisenstein
    Commented Feb 24, 2021 at 11:26
  • $\begingroup$ I have used $|x| \ge 0$ only no, I think that should be clear, Or do you mean that you need the graph of the funciton @Eisenstein $\endgroup$ Commented Feb 24, 2021 at 11:27
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    $\begingroup$ Your argument is sound, no doubt. I just requested something more general, that generalises to all such problems, not just this one. $\endgroup$
    – Eisenstein
    Commented Feb 24, 2021 at 11:29
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    $\begingroup$ Ok Will try to make one $\endgroup$ Commented Feb 24, 2021 at 11:31
  • $\begingroup$ Ok got it, no need for diagram. Also accepted the answer. $\endgroup$
    – Eisenstein
    Commented Feb 24, 2021 at 11:41
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I've changed my answer.

Your question as I see it, is like asking,

"Suppose we have a wrong equation:

$1 + 0 = 4$.

Then:

$1 + 0 + 1 = 4 + 1$

$2 = 5.$

What is wrong with the final equation?"

But perhaps I am not understanding the question?

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  • $\begingroup$ You just made me feel more stupid. But hey, the question was not absolutely dumb! Many have problems regarding this. $\endgroup$
    – Eisenstein
    Commented Feb 24, 2021 at 11:46
  • $\begingroup$ I never said the question was dumb. I even said, perhaps I don't understand the question. $\endgroup$ Commented Feb 24, 2021 at 11:47
  • $\begingroup$ Your assessment is correct, and you understood the question extremely well, and I also knew the absolute value issue from the very beginning, thought it was minor, but now I see it is integral. Haha! $\endgroup$
    – Eisenstein
    Commented Feb 24, 2021 at 11:49
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The third to fourth line, $|x-1| < \frac \epsilon 3 \implies |x-2| < \frac \epsilon 3 -1$ is dead wrong.

$|a| < M$ most certainly does NOT imply $|a-k| < M-k$.

If $k > M$ then $M -k< 0$ which would make this impossible. And as we are taking $\epsilon$ to be arbitrarily small, it will almost certainly be the case that

$|x-1| < \frac \epsilon 3 < 1$ so

$|x-1| - 1 < \frac \epsilon 3 -1 < 0$.

You seem to assume that $|x-1| - 1 = |x-1| - 1 = |x-2|$ but that would only be true if $x-1 \ge 1$. But as we are assumimg $|x-1| < \frac \epsilon 3$ that will not be true at all (even if we assume $x-1 > 0$ we don't have $x-1 \ge 1$).

(As well if $a < 0; k > 0$ then $a-k < a < 0$ and $|a-k| > |a|$. In fact $|a-k| = |a| + k$. So $|a| < M$ will not imply $|a-k| = |a| +k < M -k$.)

....

Actually the triangle inequality that says $|a| + |k| \ge |a + k|$ can be manipulated to show $|a - k|\ge |a| -|k|$

so we have $|x-2| = |x-1 -1| \ge |x-1| -1 < \frac \epsilon 3 -1$ just .... kills this.

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