Interesting limit with bijective function

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Source: Romanian District Olympiad 2025 11.3

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#1 h Mar 8, 2025, 1:09 PM • 1 Y

Y by Creativename27

Let $f:[0,\infty)\rightarrow [0,\infty)$ be a continuous and bijective function, such that $\lim_{x\rightarrow\infty}\frac{f^{-1}(f(x)/x)}{x}=1.$

Show that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\infty$ and $\lim_{x\rightarrow\infty}\frac{f^{-1}(ax)}{f^{-1}(x)}=1$ for any $a>0$ .
Give an example of function which satisfies the hypothesis.

This post has been edited 1 time. Last edited by AndreiVila, Mar 11, 2025, 10:32 AM

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#2 h Mar 11, 2025, 8:20 AM • 1 Y

Y by Creativename27

Proof for the corrected problem: 1. Since $f^{-1} ( f(x) / x ) \to \infty$ and $f^{-1}$ is bijective and thus strictly increasing, we have $f(x) / x \to \infty$ .
For the second claim, it suffices to show that $\liminf_{x \to \infty} \frac{f^{-1}(ax)}{f^{-1}(x)} \ge 1$ , then the upper bound follows by replacing $x \leftarrow a^{-1}x$ . So we need to show that for any $\delta>0$ , for large enough $x$ we have
$\frac{f^{-1}(ax)}{f^{-1}(x)} > 1-\delta \Longleftrightarrow ax > f((1-\delta)f^{-1}(x)) .$ Replacing $x \leftarrow f(x)$ , this is equivalent to $af(x) > f((1-\delta)x)$ for large $x$ . But from the condition, we know for all sufficiently large $x$ that
$f^{-1} \left( \frac{f(x)}{x} \right) > (1 - \delta) x \Longrightarrow f(x) > x f((1-\delta)x).$ Hence the assertion is true for $x>a^{-1}$ large enough.
2. An example would be $f(x)=e^x-1$ .

This post has been edited 1 time. Last edited by Gryphos, Mar 11, 2025, 10:00 AM

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#3 h Mar 11, 2025, 9:40 AM

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Gryphos wrote:

Are you sure this is the correct problem? Since $f^{-1} ( f(x) / x ) \to \infty$ , we must in fact have $f(x) / x \to \infty$ .
An example would be $f(x)=e^x-1$ .

yes, you were asked to prove that $lim_{x \to \infty} \frac{f(x)}{x} = \infty$

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#5 h Mar 17, 2025, 3:46 AM

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a). Since $f$ is continuous, it follows that:
$f^{-1} \left( \frac{f(x)}{x} \right) \to \infty \text{ as } x \to \infty.$
For all $M > 0$ , there exists $N > 0$ such that for all $x > N$ , we have $f(x) > M$ .

Thus,
$f^{-1} \left( \frac{f(x)}{x} \right) > M.$
This implies:
$f(x) > x f(M).$
Taking the limit as $x \to \infty$ , we get:
$\lim_{n \to \infty} f(x) \to \infty.$
Thus, $f(x)$ is strictly increasing.

$1 - \varepsilon < \frac{f^{-1} \left( \frac{f(x)}{x} \right)}{x} < 1 + \varepsilon.$
$f(x(1 - \varepsilon)) < \frac{f(x)}{x} < f((1 + \varepsilon)x).$
Since $f$ is increasing, it follows that:
$\frac{f(x)}{x} \to \infty.$
another cute way of proving the function is strictly increasing is

This post has been edited 1 time. Last edited by Levieee, Mar 17, 2025, 3:48 AM

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#6 h Yesterday at 12:09 PM

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Not being decreasing does not imply the function is increasing( take f(x)=x^2 on R). Also, if a function goes to infinity as x goes to infinity doesn't necessarily mean it's increasing. An example would be x(2+sin x)

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#7 h Yesterday at 12:14 PM

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TurtleKing123 wrote:

Not being decreasing does not imply the function is increasing( take f(x)=x^2 on R). Also, if a function goes to infinity as x goes to infinity doesn't necessarily mean it's increasing. An example would be x(2+sin x)

no, bijectivity is given, $x(2+sin x)$ isn't bijective also $x^{2}$ isn't bijective either

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#8 h Yesterday at 12:25 PM

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Yeah, that's the whole argument, being bijective and continuous implies monotony, but you didn't mention.

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#9 h Yesterday at 12:35 PM

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TurtleKing123 wrote:

Yeah, that's the whole argument, being bijective and continuous implies monotony, but you didn't mention.

i thought that would be obvious, i also didn't mention why f being continuous implies the first line, noone said anything, it's true yes but maybe writing it down would be better, it's in noway a 100% rigorous solution but a sketch, and it's obvious that a bijective and continuous function implies monotonicity

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