Equivalent Definition of a Continuous Function I am trying to prove the following are equivalent.
Theorem:
$$\forall \epsilon >0, \exists \delta >0: \forall x \in I, \tag{1}[|x-a|<\delta \implies |f(x)-L|<\epsilon]$$
$$\lim_{x\to a} f(x)=f(a). \tag{2}$$
The definition of a limit that I am using is $\forall \epsilon >0, \exists \delta >0: \forall x \in I, [0<|x-a|<\delta \implies |f(x)-L|<\epsilon]$.
Proof:
$\impliedby$ This way is obvious.
$\implies$ Assume $(1)$. Show $(2)$. Using the definition of a limit below two. We let $\epsilon >0$. Take $\delta >0$, and let $x\in I$. Assume $0<|x-a|<\delta$. By $(1)$, we know $|f(x)-L|<\epsilon$. And now I am trying to show $L=f(a)$. I am not sure of the best way to proceed from here.
 A: Assume that $(1)$ holds.
It is clear that if $(1)$ holds, then $\lim\limits_{x\to a}f(x) = L$. The question is, as you note, how to show that it must be the case that $L=f(a)$.
Let $\epsilon = \frac{1}{2}|f(a)-L|$. If $\epsilon\gt0$, then by $(1)$, there exists $\delta\gt 0$ such that if $|x-a|\lt\delta$, then $|f(x)-L|\lt\epsilon$. Take $x=a$; then $|x-a|=0\lt \delta$, so it must be the case that $|f(a)-L|\lt\epsilon = \frac{1}{2}|f(a)-L|$. This is impossible. Thus, it must be the case that $|f(a)-L|=0$; that is, $f(a)=L$.
Alternatively, for every $\epsilon\gt 0$, whatever $\delta$ happens to be, $|a-a|\lt \delta$, hence $|f(a)-L|\lt\epsilon$. That is,
$$\forall\epsilon\gt 0 \Bigl( |f(a)-L|\lt\epsilon\Bigr).\tag{3}$$
The only nonnegative number that can satisfy $(3)$ is $0$, hence $|f(a)-L|=0$, proving that $f(a)=L$.
A: 
And now I am trying to show L=f(a). I am not sure of the best way to proceed from here.

If condition $1$ holds then if $|x-a| = 0$ then $|x-a| < \delta$ for all $\delta$ and so then $|f(x)- L| < \epsilon$ for all $\epsilon > 0$.
But if $|x-a| = 0$ then $x = a$ and $f(x) = f(a)$.  And so we have that $|f(x) -L |=|f(a)-L| < \epsilon$.  But epsilon can be any positive number so $|f(a)-L|$ is smaller than any positive number.
There is no positive number that is smaller than all positive numbers as that would mean the number is smaller than itself.
So $|f(a) - L|$ is not positive.   But it's not negative either so it must be $0$.
So $f(a) = L$.
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First note condition 1 can only hold at all if $L = f(a)$.
If the condition is true then $|a-a| = 0 < \delta$ for all $\delta$.  So $|f(a)-L| < \epsilon$ for all $\epsilon > 0$.  So $|f(a)-L| \ge 0$ but it is strictly smaller than any positive number (including itself if it isn't $0$) so it must be $0$ and $f(a) =L$.
If condition one holds:
Then for any $\epsilon > 0$ and any $\delta$ at all we will have for $|x-a| = 0 \implies x = a\implies f(x) =f(a)$.  So we must have $|f(a) - L|=|f(x)-L| < \epsilon$ for all $\epsilon> 0$.  But $|f(a) - L|$ is a constant is is greater or equal to $0$ but it is less than any possible positive number.  So it is zero.  ANd $f(a) = L$.
So condition one is now saying that for any $\epsilon > 0$ there is a $\delta > 0$ so that $|x-a| < \delta$ whether or not $|x-a| =0$ or $|x-a| > 0$ we will have $|f(x)- f(a)| < \epsilon$.  So if $|x-a| < \delta$ and $|x-a| > 0$ we have |f(x) - f(a)| < \epsilon$.
And that is the definition of $\lim_{x\to a} f(x) = f(a)$.
......
If condition two holds.
If $\lim_{x\to a}f(x) = f(a)$ then condition one will hold but only if we set $L =f(a)$.  If $L \ne f(a)$ then for $x=a$ we will have $|x-a|=0 < \delta$ but $|f(a)-L| > 0$ so for any $\epsilon: 0 < \epsilon < |f(a)-L|$ will have condition $1$ fail.
But if we allow $L = f(a)$ then by definition of limit we have for all $\epsilon > 0$ there is $\delta > 0$ so that $0 < |x-a| < \delta \implies |f(x)-L| = |f(x)-f(a)| < \epsilon$.
But if $|x-a| = 0$ we will have $x=a$ and so $|f(x) - f(a)| = |f(x) - L| =0 < \epsilon$.
SO condition 1 holds.
