What means the notation $\lim_{x\to +\infty} f(x)<1$ for a real function $f$ with infinite singularities?

Question

Let $f$ be a real function which possesses an infinite set (of points) of singularities. I think about, for example to $$f(x)=\frac{1}{\cos(x)},$$ but just to give an example.

During the calculus class today we discussed about if it possible for a real function with an infinite set of points of singularities to be below a positive constant $c$ (to fox the ideas, we can say below $1$). Someone said that it would be enough to ask that $\lim_{x\to +\infty} f(x)<1$.

Actually I am not convinced about that because what prevents that $f$ has a singularity at infinity? I mean, about the example I wrote above, what does mean "$x\to +\infty$"? Do you have in mind any examples of real function with infinite singularities for which makes sense the writing "$x\to +\infty$" instead?

Thank you in advance.

Answer 1

Certainly it is not true that $\lim_{x \to \infty} \frac{1}{\cos(x)}$ exists; in particular the usual definition is that if the limit $L$ exists, then we must have for any $\epsilon > 0$, there is some bound $M$ such that $x > M \implies \left|\frac{1}{\cos(x)} - L\right| < \epsilon$. Since $\frac{1}{\cos(x)}$ takes on every value outside of $(-1, 1)$ on every interval $[x, x + 2\pi)$, this can't happen.

The issue is that one needs the set of singularities to be bounded; in particular consider a function like $$ f(x) = \frac{e^{-x^2}}{\sin(x^{-1})} $$

Away from $[-1, 1]$, this function is well defined, but it certainly has infinite singularities at $\pm (\pi n)^{-1}$ for $n \in \mathbb Z$. And yet we see that as $x \to \pm \infty$, $f(x) \to 0$.

Answer 2

Remember that the epsilon-delta definition of a limit at infinity is that for any $\varepsilon > 0$ we can find $M$ such that $x > M \implies |f(x) - L| < \varepsilon$. Since it has to hold for any positive $\varepsilon$, it has to hold for any value we could choose, for example if $\varepsilon = 1$ then there has to be a point where $|f(x) - 1| < 1$ for all $x > M$, which implies $0 < f(x) < 1$.

So if $f(x)$ has any essential singularities, they have to be bounded away from $+\infty$, i.e. they must all exist at points $x < M$. Which is fine - you can create a function with infinitely many singularities in the interval $(-\infty, M)$, as long as it then behaves nicely after that.