If $\epsilon$ is arbitrary, why is $\epsilon$ zero here? If Epsilon is zero as in the last line of the answer provided, then wouldn't the limit definition fail? So Epislon has to be larger than zero? Please see the last line of the proof, taken from MathOnline
 A: The proof is divided in four parts. In the first one says "let $\varepsilon>0$ be given" and the third part finishes with $|L-M| < \varepsilon$. Hence, the proof shows the following:

Proposition. For every $\varepsilon>0$, $|L-M| < \varepsilon$.

Now, how can exactly conclude that $|L-M| = 0$? Well, what happens if $|L-M| \neq 0$? Note that if $|L-M| \neq 0$ then $|L-M| > 0$, and then, taking $\varepsilon = |L-M|$ in the proposition we end up with $|L-M| < |L-M|$, a contradiction. Hence $|L-M| = 0$.
A: We never put $\epsilon=0$, but we let $\epsilon$ tend to zero.
If some quantity $Q$ satisfies $|Q|<\epsilon$ for every $\epsilon$, then by letting $\epsilon\to 0$ we conclude that $Q=0$. This is because the inequality holds for $\epsilon=\frac{1}{n}$ for every $n$ (for example), and so $|Q|<\frac{1}{n}$ for every natural number $n$. Now let $n\to\infty$. Another way to see it: if it weren't true that $Q=0$, then $|Q|>0$, and so by taking $\epsilon=\frac{|Q|}{2}$ we would violate our assumption that $|Q|<\epsilon$ for every $\epsilon>0$.
A: Epsilon is not zero but arbitrary. In the definition of the Limit it was shown that $f$ converges to $L$ if we can get $f(x)$ arbitrarily close by restricting $|x-c|< \delta$.
In the next step, we show das $L$ and $M$ must be arbitrarily close to each other because they are both arbitrarily close to $f$ for a sufficiently small $\delta$.
