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Let $f: \Re \mapsto \Re$ be continuous function. How to prove that if $\lim_{x \rightarrow \infty } |f(x)| = \infty$ then $\lim_{x \rightarrow \infty } f(x) = \infty$ or $\lim_{x \rightarrow \infty } f(x) = -\infty$ ?

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closed as off-topic by Siminore, apnorton, Najib Idrissi, Travis, Hippalectryon Jan 23 '15 at 16:05

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  • $\begingroup$ Eventually, $|f(x)|>1$; this implies, by continuity, that $f$ eventually has a constant sign. $\endgroup$ – David Mitra Jan 23 '15 at 14:57
  • $\begingroup$ Counter-example to your result. Consider f(x)=[x . sin(x)] which equals to x when sin(x)=1 and which equals to (-x) when sin(x)=-1. So this function oscillates between positive and negave numbers and its absolute value tends to + infinity. $\endgroup$ – Idris Jan 23 '15 at 23:16
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For any $M > 0$ there exists $x_0 > 0$ with the property that $x > x_0$ implies $|f(x)| > M$. In particular, $x > x_0$ implies $f(x) \not= 0$ so beyond that point $f(x)$ is either always positive or always negative. Thus $x > x_0$ implies either $f(x) > M$ or $f(x) < -M$.

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  • $\begingroup$ in this line you wrote "beyond that point f(x) is either always positive or always negative". but how you conclude that? $\endgroup$ – user2637293 Jan 23 '15 at 15:29
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    $\begingroup$ If there were points $a > x_0$ and $b > x_0$ such that $f(a) > 0$ and $f(b) < 0$, the stated hypothesis that $f$ is continuous would imply there is a point $c$ in between $a$ and $b$ (and thus $c > x_0$) for which $f(c) = 0$. This contradicts $|f(c)| > M$. $\endgroup$ – Umberto P. Jan 23 '15 at 15:47

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