From Rudin's book Principles of Mathematical Analysis Definition 4.1
For every $\epsilon>0$ there exists a $\delta>0$ such that
$$d_Y(f(x),q)<\epsilon$$
for all points $x\in E$ for which
$$0<d_X(x,p)<\delta$$
I think a limit $\lim_{x\to p}f(x)=q$ is something that we don't care about the point $p$ and $q$, we just focus on the tendency.
It makes sense in $0<d_X(x,p)<\delta$, we don't care the point $p$.
However, why it is $d_Y(f(x),q)<\epsilon$, it just take the $q$ into account. I think it should be $0<d_Y(f(x),q)<\epsilon$, so that the $q$ isn't take into account.
I know that $d_Y(f(x),q)<\epsilon$ is a bigger set of $0<d_Y(f(x),q)<\epsilon$, which means if we found some $0<d_X(x,p)<\delta$ satisfy $0<d_Y(f(x),q)<\epsilon$, it do satisfy the $d_Y(f(x),q)<\epsilon$, so the $d_Y(f(x),q)<\epsilon$ is more harder.
but why, why we choose $d_Y(f(x),q)<\epsilon$ rather than $0<d_Y(f(x),q)<\epsilon$.
Best Answer
That is true: we don't care about the exact value at point $p$ (hence the $0 < d_X(x,p) < \varepsilon$), and we don't care whether $f(x)$ becomes exactly $q$.
However, in the former case, just asking for $0 < d_X(x,p) < \delta$ is a relaxation: it literally means "there is no condition you have to verify at the specific point $p$ itself." If we added $0 < d_Y(f(x),q) < \varepsilon$ to the latter, however, it would be a strenghtening: it would add a condition (namely, "on top of the rest, you cannot take the value $q$).
That's the idea: we don't care about the value at $p$, so we don't have any constraint at $p$. And indeed we don't care whether the value becomes exactly $q$, so we don't add that constraint — but, more importantly, we also don't care that the value does not become $q$, so we don't add that extra constraint either saying "you cannot take the value $q$." That wouldn't capture what we want.