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I have been using the big O notation for some time now and know that, for example, $(e^x)^2 \in o(e^{x^2})$. In general, if $f$ grows faster than $g$, then $f(g(x))$ grows faster than $g(f(x))$. Does this hold in general? If so, how can I prove this? I was not able to find such proof anywhere. If not, does something similar hold, which would imply examples such as the one mentioned?


The version with "$f(g(x))$ grows strictly faster than $g(f(x))$" clearly does not hold (as correctly pointed out by Brian). Does the non-strict version hold?

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It definitely does not hold in general. Let $f(x)=x^2$ and $g(x)=x$; then

$$\lim_{x\to\infty}\frac{g(x)}{f(x)}=\lim_{x\to\infty}\frac{x}{x^2}=0\;,$$

so $g(x)\in o(f(x))$, but $g\circ f=f\circ g$.

Taking $f(x)=1$ and $g(x)=\frac1x$ yields a similar example.

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    $\begingroup$ +1 Thanks prof. Brian! I dont understand why this answer got two downvotes. This answer perfectly answers OPs question. In a sense, it is vandalism to downvote an answer of this quality. $\endgroup$
    – James
    Apr 10 '20 at 22:18
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    $\begingroup$ @James: You’re welcome! $\endgroup$ Apr 10 '20 at 22:20
  • $\begingroup$ @James, it does not. Namely, the part "If not, does something similar hold...", which I assume does, as I have written in the edit. While I am grateful for this answer, it does not resolve the original question and only answers the not-so-interesting part while keeping the main thing unanswered. $\endgroup$ Apr 11 '20 at 16:31
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    $\begingroup$ @user2316602: Agreed. It is, however, a correct answer to part of your question, and I have to admit that I, like James, find the downvotes a bit odd. (The other part really does seem quite non-trivial, though this is hardly my field.) $\endgroup$ Apr 11 '20 at 16:39

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