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Suppose $a\in \mathbb{R}$, $a \in (0,1)$ and a function $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying the following property:

$(1)$ $\lim_{x \rightarrow \infty}{f(x)}=0$

$(2) \lim_{x \rightarrow \infty}{\frac{f(x)-f(ax)}{x}}=0$

Show that $\lim_{x \rightarrow \infty}{\frac{f(x)}{x}}=0$

This is a problem from 'Putnam and Beyond', page $127$.

My attempt: Clearly $\lim_{x \rightarrow \infty}{f(x)}=0 \Rightarrow \lim_{x \rightarrow \infty}{f(ax)}=0 \Rightarrow \lim_{x \rightarrow \infty}{\frac{f(ax)}{x}}=0 \Rightarrow \lim_{x \rightarrow \infty}{\frac{f(x)}{x}}=0$.

Question: Does the last arrow hold? I obtain it using $(2)$ and distributivity of limit. But I don't know whether the limit $\lim_{x \rightarrow \infty}{\frac{f(x)}{x}}$ exists or not.

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    $\begingroup$ Why not $\lim\limits_{x\to\infty}\frac{f(x)}{x}=\lim\limits_{x\to\infty}\frac{f(ax)}{x}+\lim\limits_{x\to\infty}\frac{f(x)-f(ax)}{x}$ where the latter two exist for sure. ? $\endgroup$ Commented May 27, 2015 at 8:06
  • $\begingroup$ the correct question should be like math.stackexchange.com/q/568087/72031 $\endgroup$
    – Paramanand Singh
    Commented May 28, 2015 at 8:44
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    $\begingroup$ There is a typo in the book (as is so evident from the solution given there). The correct question should have limits $x \to 0$ instead of $x \to \infty$. Also $a > 0, a \neq 1$. It is not really necessary to have $a < 1$. $\endgroup$
    – Paramanand Singh
    Commented May 28, 2015 at 9:16

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I think there is something missing. It looks like obvious:

$$\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\frac{0}{\infty}=0.$$

Here i'm not using other hp.

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