What exactly goes wrong with this definition of a limit What goes wrong when I define $\displaystyle{\lim_{x \to c} f(x) = L}$ as $$ \forall \epsilon > 0, \forall \delta > 0; 0 < |x-c|<\delta \implies |f(x)-L|<\epsilon$$
 A: What goes wrong?  Everything.
Let $f(x) = x^2$ and let $c = 2$ 
Claim $\lim_{x\to 2} x^2 \ne 4$.
Pf:  Let $\epsilon = 0.1$.  Let $\delta = 7$.
So if $x=6$ then $0 < |x-2| = |6-2| = 4 < \delta$.  But $|f(x) -4|=|36-4|=32 \not < 0.1 = \epsilon$.
So $\lim_{x\to 2}x^2 \ne 4$.
Very few functions and points will have limits.  I think only constant functions will.  Or functions that are constant everywhere but the value we are taking the limit toward.
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You can't do it for ALL $\delta$.
If $x_0 \ne c$ and  $f(x_0) \ne L$ then we can take $\delta >|x-x_0|$ and $\epsilon < |f(x_0) -L|$. And then  you have .....
$|x-x_0| < \delta$ and $|f(x) - L| > \epsilon$.
So $\lim_{x\to c} f(x) \ne L$. Ever.  (Assuming $f$ has at least one point other that $c$ where $f(x) \ne L$.)
A: You should have for every epsilon, there exist a delta such as...
Else it will not work. We need to pick one delta, and show that every epsilon > 0 is less than f(x0) - L, thus their distance goes to 0, and hence f(x) -> L for sufficiently small epsilon. Letting this be true for all deltas would result in that every two point in the domain will make the limit go to L, so it would make the function a constant.
