# Epsilon-delta proof regarding existence of a limit

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 $$f(x)$$.

• 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. Mar 9, 2019 at 15:33
• Yes, the limit exists and equals 3. The $\delta$ in the $\epsilon-\delta$ proof is $\delta=\frac{\epsilon}{2}$ Mar 9, 2019 at 15:38
• 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. Mar 9, 2019 at 16:35

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?

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 $$|x - 1| < \frac \epsilon 2$$. 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 $$\lim\limits_{x\to 1} f(x)= 3$$.

....

 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 $$0 < -(x-1) < \frac \epsilon 2 \iff$$

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

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