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The question is as follows:

Suppose $D \subset \mathbb{R}$ and $f:D\rightarrow \mathbb{R}$ is monotone increasing. Let $a \in \mathbb{R}$ be a limit point of $D\cap \left ( -\infty,a \right )$. If $f$ is not bounded above, prove that $$\lim_{x\rightarrow a^{-}}f(x)=\infty.$$

This seems rather obvious. I know that $$\lim_{x\rightarrow a^{-}}f(x)=L \in \mathbb{R},$$ if given any $\epsilon>0$, $\exists \ \delta > 0$ such that if $0<a-x<\delta$ and $x \in D$, then $$\left | f(x)-L \right |<\epsilon.$$ But since $f$ is not bounded above, $$\sup_{x \in D\cap \left ( -\infty,a \right )}f(x)$$ does not exist. How can I continue from here?

Thanks in advance!

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    $\begingroup$ If you take $D = (-\pi/2,\pi/2)$, $f(x) = \tan x$, and $a = 0$, then the claim is not true. Seems like the hypotheses need to be sharpened. $\endgroup$ – Alex Ortiz Nov 2 '18 at 1:57
  • $\begingroup$ You perhaps mean that $a$ is the largest limit point of $D$. $\endgroup$ – Paramanand Singh Nov 2 '18 at 2:45
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If lim f = r in R, then exists d with for all x,
0 < |x - a| < d implies |f(x) - r| < 1.
Thus for x near a, r - 1 < f(x) < r + 1.
Since f is increasing, f is bounded above by r + 1.

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