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I have the following problem. Let $f : \mathbb{R} \to \mathbb{R}$ be continuous and strictly increasing function and let $g : \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose $\lim_{x \to 0} \frac{f(x)}{f(g(x))} = 0$. Does this imply that $\lim_{x \to 0} \frac{x}{g(x)} = 0$?

TomH
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2 Answers2

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No. Define $f(x)=\mathop{\rm sgn}(x) e^{-1/|x|}$ (with $f(0)=0$) and $g(x)=2x$.

Greg Martin
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  • Defining $f(0)=0$ does not seem to be quite enough to make $f$ neither strictly increasing not continuous at $x=0$. But you could combine it with $-f(-x)$ for $x<0$ or something like that to make it work. – String Apr 16 '15 at 22:17
  • Interesting. But, it seems that at $x=0$, the function $f$ is not strictly increasing (it's derivative converges to $0$). – TomH Apr 16 '15 at 22:26
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    @TomH: It is strictly increasing nevertheless! This is easy to see since $\operatorname e^{-1/x}>0$ for all $x>0$, so $0<x$ implies $f(0)<f(x)$. And for $x>0$ is has positive derivative, so there it is surely increasing. – String Apr 16 '15 at 22:34
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    It's strictly increasing for the same reason that $x^3$ is strictly increasing on $\Bbb R$, despite having a derivative of $0$ at one point. Since the original question does insist on having $f$ defined on $\Bbb R$ and not just on $\Bbb R\ge0$, I modified the definition of $f$ accordingly (simply by making it odd). – Greg Martin Apr 17 '15 at 02:04
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I think that I figured it out now. Suppose that the implication does not hold, i.e. $\lim_{x \to 0} \frac{f(x)}{f(g(p))} = 0$, but $\lim_{x \to 0} \frac{x}{g(x)} = c \neq 0$.

By continuity of $g$ (at $0$) and this assumption, $g(0) = \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{x}{c} = 0$. Therefore, by continuity of $f$ and $g$ at $0$, $f(g(0)) = \lim_{x \to 0} f(g(x)) = \lim_{x \to g(0) = 0} f(x) = f(0)$. This implies that $\lim_{x \to 0} \frac{f(x)}{f(g(x))} = \frac{\lim_{x \to 0} f(x)}{\lim_{x \to 0} f(g(x))} = \frac{f(0)}{f(g(0))} = 1 > 0$. This is a contradiction.

I think I'm using an extra assumption here that is violated by Greg Martin's example, but in my particular problem (it is a part of the larger proof) it works. In any case, I would appreciate if anyone could point out, what is the extra assumption in addition to continuity that I implicitly used here.

TomH
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