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This question is from Calculus by Spivak, Chapter 8 on Least Upper Bounds:

Suppose that $f$ is a function such that $f(a) \leq f(b)$ whenever $a<b$. Prove that $\lim_{x_\to a^-} f(x)$ and $\lim_{x_\to a^+} f(x)$ both exist.

Worked on it for a while but didn't get anywhere. Any help would be appreciated.

helios321
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  • @ClementC. Thanks but I still don't know how to prove that delta exists such that the limit is the supremum of the set where ${f(x):x<a}$ – helios321 Jul 29 '17 at 12:25

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Fix $a$ in the domain of $f$ Then $\left \{ f(x):x\le a \right \}$ is bounded above by $f(a)$ and so has a least upper bound, $A\le f(a).$

By definition of $A$ as the $\textit{least}$ upper bound, if $\epsilon>0$ there is a $\delta >0$ such that if $x\in (a-\delta,a), $ then $A-f(x)<\epsilon,\ $ which says precisely that $f(a^-)=A$.

Similarly, you can show that $f(a^+)=\inf \left \{ f(x):x\ge a \right \}.$

Matematleta
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    It seems I'm missing a piece of understanding about least upper bound. Why exactly is your second sentence true? In particular why does it work for any $\epsilon$? I can't link the basic definition of least upper bound to your statement. Isn't it possible that we choose $\epsilon$ too small for such conditions to hold? – helios321 Jul 31 '17 at 10:45
  • It's the definition of least upper bound. If the statement is NOT true then $A$ is easily seen not to be the least upper bound of the set. – Matematleta Jul 31 '17 at 18:05