Hardy theorem on elementary functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:08:18Z http://mathoverflow.net/feeds/question/111260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111260/hardy-theorem-on-elementary-functions Hardy theorem on elementary functions FF 2012-11-02T08:57:43Z 2012-11-03T09:30:27Z <p>Say we have two elementary functions (see <a href="http://mathworld.wolfram.com/ElementaryFunction.html" rel="nofollow">http://mathworld.wolfram.com/ElementaryFunction.html</a> 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.</p> http://mathoverflow.net/questions/111260/hardy-theorem-on-elementary-functions/111268#111268 Answer by juan for Hardy theorem on elementary functions juan 2012-11-02T10:51:29Z 2012-11-03T09:30:27Z <p>In the book by Hardy Orders of Infinity you will find the Theorem (p. 18)</p> <p>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.</p> <p>With the definition:</p> <p>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. </p>