Is the proof of the connection between continuity and the limit of a function correct? We have done the following proof:
Definition: $\ f(x)\ $ is continuous at $\ x_0\ $ if $\ \vert x-x_0\vert<\delta\implies \vert f(x)-f(x_0)\vert<\varepsilon.$
The connection:
$\lim_{x \rightarrow x_0}f(x) = f(x_0) \iff f(x)$ is continuous at $x_0$.
The proof:
So we said it like this: We know that $\lim_{x\rightarrow x_0}f(x) = f(x_0)$
$0<|x - x_0| < \delta \implies |f(x) - \lim_{x\rightarrow x_0}f(x)| < \epsilon$
$0<|x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon$
And now we see that if we put $x = x_0$ we get: $|f(x_0) - f(x_0)| < \epsilon$ , we get $0 < \epsilon$, which holds since $\epsilon$ is a positive number.
However we have done this at lectures and said it is correct, but I do not see why it would be correct, since if we look at the first assumption, that $0<|x - x_0| < \delta$, we get  $0<0 < \delta$, which does not hold.
Did the professor make a mistake to say that the delta neighbourhood needs to be strictly bigger than zero.
 A: First of all, you have left out all the quantifiers in the definitions.  The definition of "$f$ is continuous at $x_0$" should say:  For every $\epsilon>0$ there is some $\delta>0$ such that for every $x$, if $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$.  You can't leave all that stuff out.
I think you're trying to prove the left-to-right direction of what you call "the connection".  To prove this:
Assume that $\lim_{x\to x_0} f(x) = f(x_0)$.  To prove that $f$ is continuous at $x_0$, start with "Suppose $\epsilon > 0$."  Then you have to come up with a $\delta$.  Then assume that $x$ is a real number such that $|x-x_0| < \delta$.  Then you have to prove that $|f(x)-f(x_0)|<\epsilon$.
The use of "$0<|x-x_0|<\delta$" in the definition of limit is correct.  But if you follow the proof strategy suggested above, you should be able to do the proof.
Note:  You may be confusing yourself by thinking of this as an exercise in manipulating the symbol "$\Rightarrow$".  You're better off writing English.  "$\Rightarrow$" means "if ... then ...".   The simplest way to prove an if-then statement is to assume the "if" part and then prove the "then" part.  That's the strategy I suggested above.
A: So you're just trying to prove $\implies\ $ ?
That is, given $\ \varepsilon>0,\ $ you are trying to prove $\ \exists\ \delta>0\ $ such that $\vert x-x_0\vert<\delta\implies \vert f(x)-f(x_0)\vert<\varepsilon.$
And you are assuming that, given $\ \varepsilon'>0,\ \exists\ \delta'>0\ $ such that $0<\vert x-x_0\vert<\delta'\implies \vert f(x)-f(x_0)\vert<\varepsilon'.\qquad (1)$
$$$$
Ok so let $\ \varepsilon>0.\ $ Then by $\ (1)$, $\ \exists\ \delta'>0\ $ such that $0<\vert x-x_0\vert<\delta'\implies \vert f(x)-f(x_0)\vert<\varepsilon.$
So our proof is almost complete except for the point $x_0.$ For this point, we don't use $(1).$ Ok, so we ask ourselves, does the implication hold in the following statement:
for our $\delta'>0\ $ we found, $\ \vert x_0-x_0\vert<\delta'\implies \vert f(x_0)-f(x_0)\vert<\varepsilon\ $ ?
Yes, as can be seen in the logical implication truth table.
Therefore the implication in the statement we were trying to prove is true $\ \forall x\in(x_0-\delta',x_0+\delta'),\ $ and the proof is complete because we found $\ \delta = \delta'.$
