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.
 A: 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.
A: 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. 

A: Hint After applying l'Hopital's Rule a few times you'll notice that each application of the rule gives
$$C \lim_{x \to \infty} \frac{(\ln x)^k}{x} = C \lim_{x \to \infty} \frac{\frac{k (\ln x)^{k - 1}}{x}}{1} = C' \lim_{x \to \infty} \frac{(\ln x)^{k - 1}}{x} ,$$ where we have absorbed the factor of $k$ into the new constant $C'$ (and that $C$ and $C'$ are both nonzero), and that the expression in the limit is identical except for the power of $\ln x$ that appears. (Of course, the computation assumes that the power rule applies, that is, that $k \neq 0$.)
A: I wouldn't invoke L'Hôpital's rule here.
Knowing that $\lim\limits_{x\to\infty} \frac{\ln x}{x} = 0$, you get
$\lim\limits_{x\to\infty} \frac{\ln x^{1/9}}{x^{1/9}} = 0 = \lim\limits_{x\to\infty} \frac{\ln x}{x^{1/9}}$.
You can then elevate at the 9th power as $x \mapsto x^9$ is continuous to get
$$\lim_{x\to\infty} {(\ln x)^9\over x}=0$$
A: You can prove $\displaystyle \lim_{x \to \infty} (\ln x)^9 / x = 0$ without the L'Hôpital's rule.
Put $x = e^t$ then $x \to \infty \Leftrightarrow t \to \infty$ and
\begin{align}
\lim_{x \to \infty} \frac{(\ln x)^9}{x} = \lim_{t \to \infty} \frac{t^9}{e^t} = 0.
\end{align}
