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If $\lim\limits_{x\to c} f(x)$ exists, and $a \leq f(x) \leq b \:\forall x\in\mathbb{R}$, then $a\leq \lim\limits_{x\to c} f(x)\leq b$.

I was able to prove this by contradiction (i.e. assume $\lim\limits_{x\to c} f(x)=L < a$ or $L>b$, choose $\epsilon:=\frac{a-L}{2}$ in the former case), but is there a direct approach? Thanks.

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Yes: let $L=\lim_{x\to c} f(x)$. When $x$ is close enough to $c$, we have $f(x) < L+\varepsilon$; since $f(x) \ge a$ always, this shows that $a < L+\varepsilon$, or $L > a-\varepsilon$. Since this is true for every $\varepsilon>0$, we conclude that $L\ge a$. A similar argument shows that $L\le b$.

(Note, however, that we used a fact like "if $x>-\varepsilon$ for every $\varepsilon>0$ then $x\ge0$"; the standard proof of this fact is a proof by contradiction much like the one you're trying to avoid!)

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  • $\begingroup$ Seems rather obvious in hindsight. Thanks. :) $\endgroup$ – Hossmeister Feb 28 '18 at 12:34
  • $\begingroup$ Everything's more obvious in hindsight! $\endgroup$ – Greg Martin Feb 28 '18 at 18:18

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