TITLE: L'Hôpital's rule: number of iterations...
QUESTION [8 upvotes]: I have this exercise :
$$ \lim_{x\to\infty} {(\ln x)^9\over x} $$
Now, I know I must use the L'Hôpital's rule to solve this one, until I reach:
somthing*$ 1/x^2 $ ,because each time I get infinity/infinity, I am allowed to use L'Hospital's rule.
My question is: Is there any shortcut to solve such problems?
Sorry for bad English.

REPLY [3 votes]: Application of L'Hospital's Rule to $\lim_{x\to \infty}\frac{\log(x)}{x^{1/9}}$ reveals
$$\lim_{x\to \infty}\frac{\log(x)}{x^{1/9}}=\lim_{x\to \infty}\frac{\frac1x}{\frac19x^{-8/9}}=\lim_{x\to\infty}\frac9{x^{1/9}}=0$$
If $g(x)$ is the function $g(x)=x^9$, then 
$$\begin{align}
\lim_{x\to \infty}\frac{\log^9(x)}{x}&=\lim_{x\to \infty}g\left(\frac{\log(x)}{x^{1/9}}\right)\\\\
&=g\left(\lim_{x\to \infty}\frac{\log(x)}{x^{1/9}}\right)\\\\
&=g(0)\\\\
&=0
\end{align}$$

NOTE:  If $f$ and $g$ satisfy the general conditions of L'Hospital's Rule, then application of L'Hospital's Rule to the limit $\lim_{x\to \infty}\frac{f(x)}{g(x)}$, with $g(x)\to \infty$, does not require that the limit of the numerator approach $\infty$.  In fact, the limit $\lim_{x\to\infty}f(x)$ can even fail to exist.