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

Question

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.

Answer 1

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.

Answer 2

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 $k \neq 0$.)

Answer 3

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}

Answer 4

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 $$\lim_{x\to\infty} {(\ln x)^9\over x}=0$$

Answer 5

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.