Prove that a function g which diverges at $\pm\infty$ has interval [a,b] such that if $x\notin[a,b] \implies g(x) > g(0)$

Question

Let $g: \mathbb R \rightarrow \mathbb R$ be a continuous function such that $ \lim_{x\to\pm\infty} = +\infty$. Show that there exists a segment $[a,b]$ with $a < b$ such that $x\notin[a,b] \implies g(x) > g(0)$

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We prove that [a,b] exists as a direct consequence of the definition of divergence.

Because the limit at $+\infty$ is $+\infty$: $$ \exists N, \forall x\in\mathbb R, x\ge N\implies g(x) > g(0) $$

Similarly, because the limit at $-\infty$ is $+\infty$: $$ \exists M, \forall x\in\mathbb R, x\le M\implies g(x) > g(0) $$

We have thus found an interval $I = [M,N]$, with $M < N$. If $x\notin I$, then $x>N$ or $x g(0)$.

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The only part where I feel uncomfortable is stating that $M < N$. We could choose $N_0 = |N|$ and $M_0=-|M|$ and then construct the interval $I = [M_0,N_0]$. It seems to make sense to me that M > N is a contradiction, but I don't know how to show it.

Answer 1

I feel like $M>N$ is a contradiction, if you consider that $0$ must lie in one of the intervals $(-\infty, M]$, $[N, \infty)$ when there's an overlap. But, it doesn't matter for your proof, because you get to choose the values of $a$ and $b$. So, picking the $M_0$ and $N_0$ as you have works perfectly fine.