How can I show that a continuous function $f(x): \Bbb R \to \Bbb R$ has a minimum value if its limit go to $+\infty$ when $x \to \pm\infty$.

Question

The question is:

Let $f(x): \Bbb R \to \Bbb R$ be a continuous function. If $ \lim f(x)_{x \to +\infty} = \lim f(x)_{x \to -\infty} = +\infty$, then there is a point $x_0 \in \Bbb R$ such that $f$ assumes its minimum value.

What I noticed:

As $f$ is continuous, we know that $f(\Bbb R)$ is an unbounded above interval such as $(a,+\infty)$, $[a,+\infty)$ or $f(\Bbb R) = \Bbb R$. But to guarantee that it has a minimum value, we need to show that it is bounded below first and then show inf$f(\Bbb R) \in f(\Bbb R)$. So we can conclude $f(\Bbb R) = [\text{inf} f(\Bbb R), +\infty)$.

Also, I've been thinking on trying to work with the continuous function $1/f(x)$ where its limits go to zero when $x \to \pm\infty$. Is it a good Idea?

Thanks a lot.

Answer 1

We can use the definition of a limit and some arbitrary point in the image of $f$, say $f(0)$. As $$\lim_{|x|\to \infty}f(x) = +\infty$$ there exists a $M$ such that $f(x) > f(0)$ for all $|x| > M$. Then, we can consider the compact interval $[-M,M]$. Since $f$ is continuous, it takes a minimum value on a compact interval, say at $x_0$. Then we have that $$f(x) \geq f(x_0)$$ for all $x \in [-M,M]$. Additionally, we have $$f(x) > f(0) \geq f(x_0)$$ for all $x$ satisfying $|x| > M$.

As for your idea of working with $1/f(x)$, the function is not necessarily continuous: it is discontinuous at all values where $f(x) = 0$, so there is not much that can be gained using it.