Why does $\epsilon$ come first in the $\epsilon-\delta$ definition of limit? As we know, $\underset{x\rightarrow c}{\lim}f(x)=L\Leftrightarrow$ for every $\epsilon>0$ there exists $\delta>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.
My question is: why do we say that for every $\epsilon>0$ there exists $\delta>0$ and not vice versa, why not for every $\delta>0$ there exists $\epsilon>0$? 

Good question!
The answer is that if we put $\delta$ first, then our definition would no longer correspond to what we mean by the limit of a function.
Proposition Let $f$ be the constant function $0$.  Let $c,L$ be any real numbers.  Then:

For all $\delta>0$ there exists $\epsilon>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.

In other words, if we define $\lim'$ by putting the $\delta$ first, then $\lim_{x\to c}'f(x)=L$.
Proof.  Let $\delta>0$ and let $\epsilon=|L|+1$. If $0<|x-c|<\delta$ then $$|f(x)-L| = |0-L|=|L|<|L|+1=\epsilon\quad\Box$$
So our constant function $0$, which should clearly converge to $0$ at every point, under the new definition converges to every real number at every point.
In particular, limits aren't unique and the properties we expect to hold for a 'limit' no longer hold.  It just isn't a very interesting definition and it no longer captures our intuition of what limits should be.
By contrast, the definition with $\epsilon$ first exactly captures the notion of a 'limit'.
