L'Hôpital's rule: number of iterations... 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.

This is somewhat informal and verbose, but easily made rigorous.
Note that the fraction is eventually monotonically decreasing*. This means that instead of $x$ we can take any sequence $a_n$ which goes to $\infty$ as $n$ increases, and the limit will be the same as in the continuous case.
Set $a_n=e^{2^n}$, and note what happens between $\frac{(\ln a_n)^9}{a_n}$ and $\frac{(\ln a_{n+1})^9}{a_{n+1}}$:


*

*The numerator is multiplied by $2^9=512$

*The denominator is squared, i.e. multiplied with itself


It doesn't take long to reach a point where this means the entire expression is multiplied by, say, something smaller than $\frac12$, and it keeps getting multiplied by smaller and smaller numbers each time we increase $n$ by $1$. From that point on we move quickly towards $0$.
* We have
$$
\frac{d}{dx}\frac{(\ln x)^9}{x}=\frac{9(\ln x)^8\cdot\frac1x\cdot x-(\ln x)^9}{x^2}\\
=\frac{(\ln x)^8}{x^2}(9-\ln x)
$$
and we see that for all $x>e^9$, this is negative.
