Inquiry Regarding Epsilon Delta Definition of a limit This may be a stupid question but why is it that the definition of the limit is defined the way it is and not as follows 
$$\forall\epsilon>0\exists\delta>0\exists x(0<|x-a|<\delta\implies |f(x)-L|<\epsilon)$$
 A: You want the value of just "one" point $x$ to be as close to the value of the limit for each $\epsilon$. Then the limit is not uniquely defined.
For example,let $f(x)$ on $[0,1]$ be defined as $f(x) = 1$ if $x = \frac 1n$, $n \in N$ and $0$ otherwise.
What is $\lim_{x \to 0} f(x)$? From your definition, it is both $0$ and $1$. 
To see this , note that for any $\epsilon>0$, let $\delta=\epsilon$, for example, then we can find easily, some point more than $\delta$ close enough to $0$, that isn't of the form $\frac 1n$, and at this point, $f(x) = 0$, so $|f(x)-0| = 0 < \epsilon$, fitting the limit criteria.
Simiarly, given $\epsilon > 0$, let $\delta$ be smaller than $\epsilon$, and of the form $\frac 1n$. Then, if we pick $\frac \delta 2$ for example, this is more than $\delta$ close to $0$, and we have that $f(\frac \delta 2) = 1$, so that $|f(\frac \delta 2) - 1| = 0 < \epsilon$. 
Hence, both $1$ and $0$ qualify as limits of the sequence under this property. However, having two different limits is an "undesirable" property, ideally when we speak of convergence in a metric space we would like unique limits. This also impedes the notion of continuity, for if there are different limits of the same sequence, how do you compare this list to a single value of a function at a point, say. 
Hence, the notion of function limits you have prescribed is not suitable. However, full marks for at least thinking about an alternate approach.
