0
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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
    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". $\endgroup$ Nov 2, 2012 at 15:13

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.