What is wrong in this choice of $\delta$?

Question

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!

Answer 1

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.

Answer 2

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?

Answer 3

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.