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If $f$ is a uniformly continuous function from $(0,\infty) \rightarrow (0,\infty)$. Then is $\lim_ {x \rightarrow \infty} f(x+1/x)/f(x) = 1$ ?

What I have tried so far is Since $f$ is uniformly continuous we don't depend upon the choice of points.

Thus for $x$ sufficiently large $|x+1/x - x| = |1/x| < \delta$

Then by definition $|f(x+1/x) - f(x)| < \epsilon$

Which implies $|\frac{f(x+1/x)}{f(x)} - 1|/ |f(x)| <\epsilon$. But if $f$ is unbounded we have a problem.

Here I am stuck, I have a feeling that the statement in my question is false, but I am not able to prove it rigorously.

I have also tried in the sequencial method which is $|x_n - y_n|$ tends to $0$ then $|f(x_n) - f(y_n)|$ tends to $0$ by taking $x_n = x_n + 1/x_n$ and $y_n = x_n$ but in the end I am getting stuck at the same unbounded case .

If you could provide me with any hints or help I will be grateful.

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    $\begingroup$ Have you thought of analyzing two cases ($f$ is bounded and $|f|\to\infty$ as $x\to \infty$) separately? $\endgroup$ Commented Apr 8, 2020 at 4:04
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    $\begingroup$ You're making a small arithmetic error: unbounded $f$ are not an issue - $ |f(x + 1/x) - f(x)| < \varepsilon \implies |f(x+1/x)/ f(x) - 1| < \varepsilon/|f(x)| \to 0$. Similarly, if $f$ is bounded away from $0$ for large $x$ then you're fine. The problem is $f$ which decay strongly to $0$. For instance, $f = \exp(-x^2)$ serves as a counterexample. $\endgroup$ Commented Apr 8, 2020 at 4:13
  • $\begingroup$ Yes this is working, the limit is $1/e^2$. I understand my mistake thank you. $\endgroup$ Commented Apr 8, 2020 at 4:37

2 Answers 2

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The claim is false.

With $f(x)=e^{-x^2}$ you will get $$ \lim_{x\to\infty}\frac{f(x+1/x)}{f(x)}=e^{-2}. $$

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This answer offers some detail and motivation for Jyrki Lahtonen's answer.

The Mean Value Theorem says $$ \begin{align} \log\left(\frac{f\!\left(x+\tfrac1x\right)}{f(x)}\right) &=\log\circ f\!\left(x+\tfrac1x\right)-\log\circ f(x)\\ &=(\log\circ f)'(\xi)\,\frac1x \end{align} $$ for some $\xi$ between $x$ and $x+\frac1x$.

If we set $(\log\circ f)'(x)\,\frac1x=a$, we get $f(x)=e^{ax^2/2}$. Then $$ \lim_{x\to\infty}\frac{f\!\left(x+\frac1x\right)}{f(x)}=e^a $$ Note that $e^{ax^2/2}$ is uniformly continuous on $(0,\infty)$ if $a\le0$.

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  • $\begingroup$ Thank you all for taking your time to go through my problem. $\endgroup$ Commented Apr 9, 2020 at 7:29

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