0
$\begingroup$

Suppose that for any $\epsilon >0$ that the solution to $|f(x)-3|<\epsilon$ is:

$x \in (1- \dfrac{\epsilon}{2}, 1 + \dfrac{\epsilon}{2})$.

Does $\lim_{x\to 1} f(x)$ exist? If so what is its value?

I have no idea on how to formally construct the $\epsilon - \delta$ proof to prove/disprove the existence of the limit of $f(x)$.

$\endgroup$
3
  • $\begingroup$ You formally construct the proof the same way you always do. The information you're given about $f$ isn't a functional expression, but that doesn't change the overall shape of an $\epsilon$-$\delta$ proof. $\endgroup$
    – Arthur
    Mar 9, 2019 at 15:33
  • $\begingroup$ Yes, the limit exists and equals 3. The $\delta$ in the $\epsilon-\delta$ proof is $\delta=\frac{\epsilon}{2}$ $\endgroup$
    – Fakemistake
    Mar 9, 2019 at 15:38
  • $\begingroup$ You don't need to prove anything. This statement is simply saying "$\lim_{x\to 1}f(x) = 3$ but in different words. Just like if I said: "$S$ is the set of all $(x,y)$ points the distances between each $(x,y)$ and $(2,3)$ is same distance" then you'd know $S$ is a circle. There's nothing to prove; it's just another way of stating the definition of a circle. $\endgroup$
    – fleablood
    Mar 9, 2019 at 16:35

2 Answers 2

Reset to default
2
$\begingroup$

The solution $$x \in \left(1- \dfrac{\epsilon}{2}, 1 + \dfrac{\epsilon}{2}\right)$$ is equivalent with $$|x-1| < \frac{\epsilon}{2}$$ Recall that the definition says

...there exists $\delta>0$ such that...

In your case, you can take delta to be $\delta = \epsilon/2$. Now you have that for all $\epsilon>0$ there exists $\delta=\frac{\epsilon}{2}$ such that

$$|x-1|<\delta \implies |f(x)-3| < \epsilon$$

What does this say about the limit?

$\endgroup$
1
$\begingroup$

It strikes me that the very statement is simply paraphrasing via the definition that $\lim\limits_{x\to 1} f(x)= 3$ and there is nothing really to prove.

Note: $x\in (1 -\frac \epsilon 2, 1 + \frac \epsilon 2)$ is the exact same thing as saying $|x - 1| < \frac \epsilon 2$[1]. So if we set $\delta = \frac \epsilon 2$, we can put this into the "magic incantation"

For any $\epsilon > 0$, then there exists a $\delta$ (namely $ \frac \epsilon2$) so that whenever $|x - 1|< \delta$ we have $|f(x) - 3| < \epsilon$.

So $\lim\limits_{x\to 1} f(x)= 3$.

....

[1] For all $\epsilon$

$x \in (1 - \frac \epsilon 2, 1+\frac \epsilon 2) \iff$

$1- \frac \epsilon 2 < x < 1 + \frac \epsilon 2 \iff$

$-\frac \epsilon 2 < x -1 < \frac \epsilon 2 \iff$

If $x-1 \ge 0$ then $0 \le x -1 < \frac \epsilon 2$; and if $x-1 < 0$ then $0 < -(x-1) < \frac \epsilon 2 \iff$

$0 \le |x-1| <\frac \epsilon 2 \iff$

$|x-1| < \frac \epsilon 2$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .