Do you pick a delta or do you pick an epsilon beginner confusion 
Let $f : \mathbb R \to\mathbb R$ be a continuous function such that $\lim_{x\to0} f(x) = 2.$ Prove that there
exists a $δ > 0$ such that, on the interval $(−δ, δ)$, the function f is bounded.

Hello I am a beginner in college mathematics, here is my attempt.
I know that for every $ε>0$, there exist a $δ>0$ such that $|x-a| < δ$ implies that $|f(x) - L| < ε$
$|x| < δ$ = $- δ < x < δ$
$|f(x) - 2| < ε$ Then use the 2nd triangle inequality?? (Sorry I am just trying to create something)
$|f(x)| - |2| < ε$  =
$|f(x)| < ε$ +  |2| =
$-ε- 2<f(x) < ε$ + 2
So here what do I do ? Do I pick a ε? So let $ε = 1$ and now the function f is bounded? Do I pick the delta or epsilon first? Thanks -Alice
 A: Your attempt seems correct, I will just fix it a bit.
We want to prove that $f(x)$ is bounded in a small environment of $x$. The limit of $f(x)$ at $0$ is $2$. So, given every $\varepsilon>0$ one can find $\delta>0$ s.t $|x|<\delta$ implies $|f(x)-2|<\varepsilon$. So taking $\varepsilon=1$, there exists such $\delta>0$ that for every $x\in(-\delta,\delta)$, $|f(x)-2|<\varepsilon$ which is the same as saying $2-\varepsilon<f(x)<\varepsilon+2$ (no need for the second triangle inequaility which works the other way btw). So this shows $f$ is bounded in the interval.
A: I find it easier than the 2nd triangle inequality to do this.
Basic in/equalities:  $|x- 0| < \delta\iff -\delta < x < \delta \iff x \in (-\delta, \delta)$.  (But $0 < |x-0| < \delta$ implies $x\in (-\delta, \delta)$ and $x \ne 0$).
$|f(x) - 2| < \epsilon \iff -\epsilon < f(x) - 2 < \epsilon \iff 2 -\epsilon < f(x) < 2 + \epsilon \iff f(x) = (2-\epsilon, 2 + \epsilon)$.
....
Now let's think about definitions:
Def:  $f(x)$ is bounded on the interval $(a,b)$ if there is an $M\in \mathbb R$ so that for every $x \in (a,b)$ we have $|f(x)| < M$.  (Or in other words $-M < f(x) < M$).
Def: $\lim_{x\to a} f(x) = L$ means.... for every $\epsilon > 0$ we can find a $\delta > 0$ so the whenever $0 < |x-a| < \delta$ (or in other words if $x \in (a-\delta, a+ \delta)$ but $x \ne a$) then we will have $|f(x) -L | < \epsilon$ (or in other words we will have $f(x) \in (L-\epsilon, L + \epsilon)$.
.....
Now notice if $f(x) \in (L-\epsilon, L + \epsilon)$ means that $f(x)$ is bounded by $\max (|L + \epsilon| , |L - \epsilon|)$.
[That is to say:
$-\max (|L + \epsilon|, |L-\epsilon|) \le  \\
-(L -\epsilon) \le \\   
L - \epsilon < f(x) < L+\epsilon \\
\le |L + \epsilon| \le\\
\max (|L + \epsilon|, |L-\epsilon|) $
so $|f(x)| < \max (|L + \epsilon|, |L-\epsilon|) $.]
....
In terms of what we are asked to do
For any $\epsilon > 0$ we can bind $f(x) < (2 -\epsilon, 2+\epsilon)$ so that would mean $|f(x)| < 2+ \epsilon$[*]  and $f(x)$ is bounded by $2 + \epsilon$ by finding the right $\delta$.
And we can pick $\epsilon$ to be anything.
If we use your choice of $\epsilon = 1$ we can find a $\delta$ so that $x \in (-\delta, \delta) \implies |f(x) -2| < 1\implies -1 < f(x) -2 < 1 \implies 1 < f(x) < 3\implies -3 < 1 < f(x) < 3\implies |f(x)| < 3$, so $f(x)$ is bounded by $3$ in that interval.
That's good enough.
We can (but don't have to) bind $f(x)$ to within any value slightly larger than $2$ but taking a small enough $\delta$ based upon the $\epsilon$ value we choose.
And we don't have pick any epsilon.
We could say:
For and $M > 2$ we can bound $|f(x)| < M$ but letting $\epsilon = M -2$ and letting $\delta$ be so that $|x-0| < \delta \implies |f(x) -2| < \epsilon$.
That would give us:
$|f(x) - 2| < \epsilon$
$2 - \epsilon < f(x) < 2 + \epsilon$
$-M = -2 - \epsilon < 2 - \epsilon < f(x) < 2+ \epsilon = M$ so
$|f(x)| < M$.
========
[*]  Note:  $f(x) \in (2-\epsilon, 2 + \epsilon)$ is a stronger statement than $|f(x)| < 2+ \epsilon$ but because
$-2 - \epsilon < $
$2 + \epsilon < f(x) < 2 + \epsilon$
We do have the weaker result as well.
