What's wrong with finding $\delta$ for limit at $a$, then taking $bLet us say we found a $\delta$ for some limit $f(x)$ as $x$ goes to $a = l$.
So we have $ |x - a| < \delta(\epsilon) $,
but it is also true that $|x-b| <\delta $ for some $b$ less than $a$ but greater than $0$.
Now I'll claim that the limit as $x$ tends to $b$ is $l$. What's wrong with what I did right now? For every $\epsilon$ this same $\delta$ would work. 
 A: I found your question somewhat difficult to understand, so please let me know if my response helps. The reason this doesn't work is basically entirely encapsulated by what $\lvert x-x_0\rvert$ really is. $\lvert x-x_0\rvert$ denotes the distance from $x$ to $x_0$ on the real line. When we discuss continuity of a function $f:\mathbf{R}\to\mathbf{R}$ at a point $a\in \mathbf{R},$ we are interested in understanding what happens to $f(x)$ as $x$ gets very close to $a$. Unless we are in a special case where the function $f$ behaves in some sort of global pattern, the behavior of $f$ near $a$ will not tell us anything about the behavior of $f$ near $b$, if $a\ne b$.
The idea with continuity is that given $\epsilon>0$, we can find a $\delta$ quite small so that $\lvert x-a\rvert<\delta$ implies that $\lvert f(x)-L\rvert<\epsilon$, where $L=\lim_{x\to a}f(x)$. 
However, different pairs $(a,\epsilon)$ and $(b,\epsilon)$ may demand different $\delta$'s. For instance, given the function $f(x)=x^2$, let $a=0, b=1, \epsilon=1$. Now, we know that $f$ is continuous, so in particular it is continuous at $a$ and $b$. At $a=0$, if we want $\lvert f(x)-f(0)\rvert=\lvert f(x)\rvert<1$ we can choose $\delta=1$ and stipulate that $\lvert x-0\rvert<\delta=1$. On the other hand, at $b=1$, $\delta=1$ will not work, because $\lvert x-1\rvert<1$ does not imply that $\lvert f(x)-f(1)\rvert<1$ since if $x=1.5$, $\lvert 1.5-1\rvert<1$ but $\lvert f(1.5)-f(1)\rvert=\lvert 2.25-1\rvert=1.25>1$. As such, we can see that we need to choose a smaller $\delta$. $\delta=.1$ will work, for instance.
A: $|x-a|<\delta \iff a-\delta <x<a+\delta$ and for the limit that means this inequality $|f(x)-\ell|<\varepsilon$ is satisfied $\forall x \in(a-\delta,a+\delta)$
or the suffiscent condition to satisfy $|f(x)-\ell|<\varepsilon$ is $x \in(a-\delta,a+\delta)$
$|x-b|<\delta \iff b-\delta <x<b+\delta$ and for $x \in(b-\delta,b+\delta)\implies \left\lbrace\begin{array}l |f(x)-\ell|<\varepsilon\\\text{or} \\ |f(x)-\ell|\geq\varepsilon \end{array} \right.$
A reminder below :

A: This statement

So we have $ |x - a| < \delta(\epsilon) $,
  but it is also true that $|x-b| <\delta $ for some $b$ less than $a$ but greater than $0$.

is false for some fixed $x$. Just take, by example, $a=1$, $b=1/2$, $x=2$ and $\delta=1.1$.
If we assume that $x$ is not fixed then the (order) relation between the sets $$A:=\{x\in\Bbb R:|x-a|<\delta\},\quad B:=\{x\in\Bbb R:|x-b|<\delta\}$$
can be arbitrary, that is for suitable $a$ and $b$ is possible that $A\subset B$ or $B\subset A$ or $A\cap B=\emptyset$, etc...
A: "So we have |x−a|<δ(ϵ), but it is also true that |x−b|<δ for some b less than a but greater than 0."
But that isn't true.
Just take any $x$ so that $b + \delta < x < a+\delta$.  Then $|x-b| > \delta$.
But $|x - a|$ can very easily be less than $\delta$.
Let $\max(b+ \delta, a - \delta) < x < a+\delta$ and we have both $|x -b| > \delta$ and $|x-a| < \delta$.
=====
(Actually if $b + \delta < a-\delta$ than $|x -a| < \delta \implies |x-b| > \delta$ always, so we didn't need to speculate $\max(b+\delta,a-\delta)< x< a+\delta$.  Simply stating $b + \delta < x < a+\delta$ would have been enough.)
