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 $|f(x)-0|\lt\epsilon$ $\endgroup$ – J. W. Tanner May 23 '19 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 $0 < m \leq f(x)$. $\endgroup$ – Jack Crawford May 23 '19 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.