# 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

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 $$0, we don't care the point $$p$$.

However, why it is $$d_Y(f(x),q)<\epsilon$$, it just take the $$q$$ into account. I think it should be $$0, so that the $$q$$ isn't take into account.

I know that $$d_Y(f(x),q)<\epsilon$$ is a bigger set of $$0, which means if we found some $$0 satisfy $$0, it do satisfy the $$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.

• It could be that $f(x)=f(p)$ even if $x\ne p$ (for example, a constant function) Commented Jun 11, 2019 at 17:42
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. Commented Jul 5, 2019 at 5:11

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.

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 $$x$$?

"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"
"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 $$0 would make $$(1)$$ false.
Consider, for instance,$$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.