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Say we have two elementary functions (see http://mathworld.wolfram.com/ElementaryFunction.html for the definition) $f_1,f_2\colon [0,\infty)\mapsto \mathbb{R}$ such that $\lim\limits_{x\to\infty}f_1(x)=\lim\limits_{x\to\infty}f_2(x)=\infty$. Can we say something about the existence of $\lim\limits_{x\to\infty}\frac{f_1(x)}{f_2(x)}$? I heard that some kind of answer gives theorem proved in 1930's by GH Hardy but I couldn't find it.

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    $\begingroup$ That theorem relates to the "Hardy field" (See en.wikipedia.org/wiki/Hardy_field), not to elementary functions in the sense of Liouville. The latter have no such property: just take $f_1=x(2+\sin x)$ and $f_2=x$. $\endgroup$
    – fedja
    Commented Nov 2, 2012 at 10:49
  • $\begingroup$ Yes. Note that even the muddled definition on the mathworld web site says that trigonometric functions are included among "elementary functions". $\endgroup$ Commented Nov 2, 2012 at 15:13

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In the book by Hardy Orders of Infinity you will find the Theorem (p. 18)

Any L-function is ultimately continuous, of constant sign, and monotonic, and, as $x\to\infty$, tends to $\infty$, or to zero or to some other definite limit. Further, if $f$ and $\phi$ are L-functions, one or other of the relations $f\succ\phi$, $f\asymp\phi$, $f\prec\phi$ holds between them.

With the definition:

We define a logarithmic-exponential function (shortly, an L-function) as a real one valued function defined, for all values of $x$ greater than some definite value, by a finite combination of the ordinary algebraical symbols (viz. $+$, $-$, $\times$, $\div$, $\root n \of \cdot$ ) and the functional symbols $\log(\cdots)$ and $e^{(\dots)}$, operating on the variable $x$ and on real constants.

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