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When proving if $f:\Bbb R \rightarrow \Bbb R$ has limits as $x \rightarrow \pm \infty$, then $f$ is bounded on $\Bbb R \setminus [A,B]$ for some $A,B \in \Bbb R$, should we prove that

  1. $f$ has a lower bound on $(- \infty,A)$ and an upper bound on $(B, \infty)$ or

  2. $f$ has an upper bound and a lower bound on $(- \infty,A)$ and an upper bound and a lower bound on $(B, \infty)$ ?

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    Those are the same. You should prove there exists $M > 0$ such that $|f(x)| \leq M$ for all $x \in \mathbb{R} \setminus [A,B]$. – 2017-02-13
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    Those two are equivalent, so whichever one seems easiest to you with your specific function is the one you should go for. – 2017-02-13
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    I'm sorry, I made a mistake, I edited my question – 2017-02-13

2 Answers 2

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We have: $$ \mathbb{R}/[A,B]=(-\infty,A) \cup(B,+\infty) $$

so we have to prove that there is an upper and a lower bound of the function in $I=(-\infty,A) \cup(B,+\infty)$. But , if $A\ne B$ this set has two disjoint components, so we need an upper and lower bound on each component, and the upper bound in $I$ is the greatest of the two upper bounds ( and the lower bound is the lowest of the lower bounds)

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Option 2 is the right one.

Here is a example showing that option 1 is not enough. Consider the map :

$$f:\mathbb{R}\to\mathbb{R},\,x\mapsto\cases{\frac{x}{1-x^2}\quad\mathrm{if}\,x\not\in\{-1,1\}\cr 0\quad\mathrm{otherwise}}$$

With $A=-1$ and $B=1$, we see that $f$ has finite limits at $\pm\infty$ (both are $0$), that $f$ has a lower bound $(= 0)$ on $(-\infty,A)$ and an upper bound $(= 0)$ on $(B,+\infty)$ but is not bounded on $\mathbb{R}-[-A,B]$

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