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I am trying to prove the following question:

Suppose $f : \mathbb{R} → \mathbb{R}$ satisfies $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} f(x) = -\infty$. Prove that there exists a number $\beta$ such that for all $\epsilon > 0$ there exists an $r \in (0, \epsilon)$ such that $f(\beta − r) \leq 0 ≤ f(β + r)$.

What I have right now:

If the function $f$ is continuous, then it is easy by using IVT. If $f$ is not continuous, I was thinking about setting $$ \beta = \sup\{x: f(x) \leq 0\}. $$

But this does not seem to work since if $$ f(x) = x, \ x \in (-\infty, 1) \cup (1, \infty) $$ and $$ f(1) = -1, $$ then $\beta = 1$ and there is a neighborhood of $\beta$ where the required condition is not satisfied.

Can anybody think of another route? Thanks a lot!

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Let $X = \{x : \exists b > a > x: \forall y \in (a; b): f(x) \leqslant 0\}$ - points that have an entire interval of non-positive values of $f$ in right of them. Let $\beta = \sup X$. Note that $\beta \notin X$ (otherwise we would also have $\frac{\beta + a}{2} \in X$ with corresponding $a$).

Let us take some $\varepsilon$. Take $x \in (\beta - \varepsilon; \beta) \cap X$ and then $b > a > x$ such that $f$ is non-positive on $(a; b)$. We have $b < \beta$, otherwise $\frac{\beta + b}{2}$ would be in $X$.

Now, as $\beta \notin X$, there is some $y \in (2\beta - b; 2\beta - a)$ [segment symmetric to $(a; b)$ with $\beta$ as center of symmetry] such that $f(y) > 0$. Then take $r = y - \beta$, and we have $\beta - r \in (a, b)$, $\beta + r = y$, so $f(\beta - r) \leqslant 0$ and $f(\beta + r) > 0$.

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