I'm trying to understand the difference between $0<m<f(x)\,, \forall x$ versus $0<f(x)\,, \forall x$.

My best current guess is the following:

  • $0<m\le f(x)\,, \forall x$ 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 $\lim_{x \to \infty} f(x) =0$

  • $0<f(x)\,, \forall x$ 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 $\lim_{x \to \infty} f(x) =0$.

  • $\begingroup$ Yes; if you choose $\epsilon=m/2$ there’s no $X$ past which $|f(x)-0|\lt\epsilon$ $\endgroup$ May 23, 2019 at 0:51
  • $\begingroup$ 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 $0 < m \leq f(x)$. $\endgroup$ May 23, 2019 at 1:02

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .