A detail of the definition of limit 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 $0<d_Y(f(x),q)<\epsilon$.
 A: 
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.

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.
A: Consider $f(x)=0$, a constant function on $\mathbb R$.  
Wouldn't we want $\lim_{x\to0}f(x)=0,$ even though $d_Y(f(x),f(0))=0$ for all $x$?
A: Your sentence
"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"
misses the point: This sentence should read
"When considering the limit $\lim_{x\to p}f(x)$ we don't care about the possible value of $f$ at the point $p$, we just focus on the tendency."
It could very well happen that $f$ assumes the value $q$ in some points  $x$ near $p$. The essential demand is that $f(x)$ is "near" $q$ while $x$ is sufficiently near $p$, but $\ne p$. When the point $f(x)$ is actually $=q$ then it surely is "near" $q$. Consider this example:
$$\lim_{x\to\infty}{\sin x\over x}=0\ .\tag{1}$$
Here $f(x)=0$ in many points $x$ "near" $\infty$. Your condition $0<d\bigl(f(x),q\bigr)<\epsilon$ would make $(1)$ false.
A: Consider, for instance,$$f(x)=\begin{cases}x&\text{ if }x\in\mathbb Q\\0&\text{ if }x\notin\mathbb Q.\end{cases}$$I suppose that you agree that, when $x$ is close to $0$; then $f(x)$ is close to $0$ too. In other words, $\lim_{x\to0}f(x)=0$. But if, in the definition of limit, we had $0<\bigl(d(f(x),q\bigr)<\varepsilon$, then the limit $\lim_{x\to0}f(x)$ would not exist.
