Assume that N is given and that $p_N$ is not the zero polynomial. I want to show that:


I am not able to do this direct and easy with l'Hôpital, are you able to see how to do it?

I am able to do it indirectly with l'Hôpital, can you please check if it is a valid proof logically?


From l'Hôpital we do know that:

$\displaystyle\lim_{x\rightarrow\infty}\left[\frac{e^x}{a_0+ax+\dots+a_Nx^N}\right]=\infty$,   (1)

Since the denominator clearly is not zero we have that:

$\dfrac{\lim\limits_{x\rightarrow\infty} e^x}{\lim\limits_{x\rightarrow\infty}(a_0+ax+\dots+a_Nx^N)}=\infty$,  (2)

Now assume for contradiction that:

$\displaystyle\lim_{x\rightarrow\infty}[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]\ne\infty$,  (3)

The limit cannot be a number $K$, because that would contradict (2). If the limit was $-\infty$, then the polynomial would after a while be larger than $e^x$ this also contradicts (2). If the limit do not exist the expression fluctuates, it means that the values the expression takes have to be bounded, this means that:

$[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]\le K$, for all $x \ge 0$, but this also contradicts (2).

Do you see how much work it is? Is there any way to simplify this?, it is not elegant to check the three special cases.


An idea without series:


Apply now arithmetic of limits, taking into account that $\;\frac{e^x}{x^n}\xrightarrow[x\to\infty]{}\infty\;$

  • 1
    $\begingroup$ Very nice!, thanks. $\endgroup$ – user119615 Mar 8 '14 at 12:58

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