# Difference between $0<m<f(x)\,, \forall x$ versus $0<f(x)\,, \forall x$?

I'm trying to understand the difference between $$0 versus $$0.

My best current guess is the following:

• $$0 implies that there is always a gap between $$f(x)$$ and $$0$$, such that even in the limit, $$f(x) \ne 0$$. E.g. it would be impossible for $$\lim_{x \to \infty} f(x) =0$$

• $$0 implies it is possible for $$\lim_{x \to \infty} f(x) =0$$

Is the above interpretation correct?

If so, then I'm still having difficulty understanding why, if $$f(x)=m$$ it is still impossible for $$\lim_{x \to \infty} f(x) =0$$.

• Yes; if you choose $\epsilon=m/2$ there’s no $X$ past which $|f(x)-0|\lt\epsilon$ May 23, 2019 at 0:51
• As for your last question, if you have that $f(x) = m$ for all $x$, clearly $\lim_{x\to \infty} f(x) = m$, and you chose $m>0$, so that's why you can't have $\lim_{x\to\infty} f(x) = 0$. Likewise if you have $f(x) > m$ for all $x$, the limit of $f(x)$, if it exists, must be a least $\lim_{x\to\infty} f(x) \geq m$. It's just that $f(x)$ is bounded below by a constant (greater than 0). Consider the function $f(x) = x^2 + m$, for example, where clearly $0 < m \leq f(x)$. May 23, 2019 at 1:02

You're right: it's different even if $$f$$ is continuous. For example, the function $$f:(0,\infty)\rightarrow \mathbb{R}$$ given by $$f(x)=1/x$$ is continuous, yet $$\lim_{x\rightarrow \infty}f(x)=0$$.
On the other hand, if $$f:I\rightarrow\mathbb{R}$$ is continuous and positive at some point $$a$$, then in fact $$f$$ is positive on some open interval containing $$a$$. Indeed, the set $$U:=f^{-1}((0,\infty))=\{x\in I : f(x)>0\}$$ is open because $$f$$ is continuous, and $$a\in U$$: thus there is some open interval $$J$$ such that $$a\in J\subseteq U$$. $$f(x)>0 \quad \text{ for all } x\in J.$$ Shrinking $$J$$ so that $$J$$ is closed+bounded and $$J\subseteq U$$, in fact this means $$f$$ is a positive function on a compact set, so it achieves its minimum. Thus there is some lower bound $$M>0$$ so that $$f(x)\geq M>0$$ for all $$x\in J$$.
It really all depends on which kinds of functions $$f$$ you are referring to in your OP.