Let's say, as an example:
$$f(x)=\frac{x^2 -25}{x-5}$$
for $ 0<|x-5|<\delta$
$$f(x) = x+5$$ as $x->5$
then we should say, $f(x) -> 10$ , not $f(x)=10$
If $f(x)$ really only approaches $10$, then why in the epsilon-delta definition of the limit, we prove that
$$|f(x)-10|<\epsilon$$
Doesn't this mean that $f(x)$ equals $10$ for $ 0<|x-5|<\delta$ ?
Even though $f(x)$ never actually reaches $10$ ?
In short, how is the statement $|f(x)-10|<\epsilon$ and $f(x) -> 10$ true at the same time?
Let's recall what the $\epsilon-\delta$ definition of a limit really says.
Suppose that $f$ is a function defined on a deleted neighbourhood of $x_0$. Then:
$$\lim_{x \to x_0} f(x) = L \iff \forall \epsilon > 0: \exists \delta > 0: 0 < |x-x_0| < \delta \implies |f(x) - L| < \epsilon$$
Notice that $f(x)$ does not have to be defined at $x_0$ itself. It has to just be defined on any deleted neighbourhood of $x_0$.
The point of the definition is that the value of the function at the point has no bearing whatsoever on the limit of the function at the point.
Now, if you've noticed, $|f(x) - L|$ really is the distance between $f(x)$ for some $x$ in the deleted neighbourhood and the real value $L$, which is supposedly the limit.
So, the definition really is just saying that if you want the distance between $f(x)$ and $L$ to be less than $\epsilon$, you need to make sure that the distance between $x$ and $x_0$ is less than $\delta$, where $\delta$ is something you choose. Of course, the distance between $x$ and $x_0$ is formally expressed as $|x-x_0|$.
In your specific case, you're showing that you can restrict the distance between $f(x)$ and $10$ by restricting the distance between $x$ and $5$. You must realize that $f(x)$ is NOT actually defined at $x = 5$ due to a division by 0.
I hope this assists you in understanding the definition. Please do ask more questions (or hit me in the head if I've misinterpreted what you were asking).