Epsilon delta with point discontinuity?

Question

Please forgive me if this question has a trivial answer, but I’m a bit stuck after coming up with this scenario. If there is a function with a point/removable discontinuity, for example $f(x)=1$ when $x\neq 0$ but $f(x)=0$ when $x=0$. The limit as $x$ approaches $0$ is $1$ but with the epsilon delta definition, if we take $\epsilon = 0.1$, then if $x=0$ (which is in the range $|x-a|<\delta$), $|f(x)-1|>0.1$, which is the negation of the epsilon delta statement. What exactly am I missing?

Answer 1

The $\epsilon-\delta$ definition for continuity is different form the $\epsilon-\delta$ definition for limit, notably for continuity the value $x=x_0$ is allowed but not for the limit for which $x\neq x_0$.

Answer 2

The original question was posted long time ago. However, I had the same doubt. I think I understand it a little better after some research.

So the Epsilon-delta definition states we can find a delta around x to satisfy that if |x-a|< delta then |f(x)-L|< any small number (Epsilon). Here f(x) doesn't necessarily need to be evaluated at "a". The definition such state that f(x) evaluated at all values at |x-a|< delta. Meaning delta is always >0 and cannot be equal to zero.

here's a nice video explaining that: https://www.youtube.com/watch?v=0sCttufU-jQ