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Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$? If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.

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    Is this a homework problem? What have you tried so far?2012-05-13
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    No. So far I am convinced that the answer to both questions is yes and tried to prove the second one. Suppose that $\exists x_0\in (a,b):f(x_0)<\ell$. Then, $\exists \xi>0:a$\epsilon=f(x_0+\xi)-f(x_0)$ to get a contradiction but I can't show that. – 2012-05-13
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    I have reached this point: $\exists \delta>0:0x_0-\xi$ I am done.2012-05-13

2 Answers 2

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Both statements are true.

There are two cases to consider. Let $S=\{f(x):x\in(a,b)\}$. Suppose first that $S$ is bounded below, i.e., that there is some $y\in\Bbb R$ such that $y

To prove this, let $\epsilon>0$. Then $u+\epsilon>u$, so $u+\epsilon$ is not a lower bound of $S$, and there is therefore some $a+\delta\in(a,b)$ such that $f(a+\delta)

The second is an immediate consequence of the argument just given: $u\le f(x)$ for every $x\in(a,b)$.

Now suppose that $S$ is not bounded below. Then for every $u\in\Bbb R$ there is an $x_u\in(a,b)$ such that $f(x_u)

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    You say that $u=\inf S$ and then that $u$ is not a lower bound of $S$. Do you mean $u+\epsilon$?2012-05-13
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    @SomeoneContinuous: Yes, that was a typo. I also still have to add the other case.2012-05-13
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    At the second line, didn't you mean $u\in \mathbb{R}$?2014-07-15
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    Actually it follows that $l=\lim_{x\to a} f(x)$f$ is strictly increasing on the open interval $(a,b).$2015-12-05
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    @azc: True, but I was addressing the actual question.2015-12-05
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As for the second statement, I think there is a more precise result. Below is my trial, please check if it is correct.

Lemma Let $f\colon D\subset\mathbb{R}\to\mathbb{R}.$ Suppose that $a, b\in D$ with $af(a+0),\qquad\forall x\in (a,b)\cap D. \end{gather*}

Proof: Let $A=f(a+0):=\lim\limits_{x\to a+0}f(x).$
If $A=-\infty,$ then there is nothing to show. So we assume that $A\in\mathbb{R}.$

We prove the statement by contradiction. If otherwise there exists $x_1\in (a,b)\cap D$ such that $f(x_1)0$ with $\deltaA-\epsilon_1=A-\frac{A-f(x_1)}{2}=\frac{A+f(x_1)}{2}>f(x_1). \end{gather*} Pick $y_1\in (a,a+\delta_1),$ then $af(x_1).$ But this contradicts that $f$ is strictly increasing on $(a,b)\cap D.$ If otherwise there also exists $x_2\in (a,b)\cap D$ such that $f(x_2)=A,$ then, pick arbitrarily $x_3\in (a,x_2),$ then, by strict increasingness of $f$ on $(a,b)\cap D,$ we have $f(x_3)f(a+0),$ for all $x\in (a,b)\cap D.$