$\lim\limits_{x \to \infty}\frac{\ln^{1000} x}{x^5}$

I'm trying to solve $$\lim\limits_{x \to \infty}\frac{\ln^{1000} x}{x^5}$$ Here's what I get: $$e^{\lim\limits_{x \to \infty}\ln{\frac{\ln^{1000}x}{x^5} }}$$ Dropping the $e$ for ease, $$\lim\limits_{x \to \infty} 1000\ln{(\ln{(x)})} - 5 \ln{x}$$

Now I have $\infty - \infty$.. I know there must be a next step, but I don't know what it would be.

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Is $\ln^{1000}x$ intended to mean $(\ln x)^{1000}$? That's usual withtrigonometric functions, but I had never seen it used with logarithms... – Mariano Suárez-Alvarez♦ Feb 9 '11 at 22:03
The limit after "Here's what I get" it not at all the same as what you started with. – Hans Lundmark Feb 10 '11 at 7:28
@Hans: Why?  It was writing $e^{ln{y}}=y$ – bobobobo Feb 13 '11 at 18:26
Sorry, that was just me doing some sloppy reading! I thought it said "limit, as $e^x \to \infty$, of ..." instead of "e to the limit of ...". – Hans Lundmark Feb 13 '11 at 18:57

3 Answers

HINT $\$ Changing variables $\rm\ Z = ln\ X\$ yields $\displaystyle\rm\ \lim_{\ Z\ \to\ \infty}\ \frac{Z^{1000}}{e^{5\:Z}}\$ which is easily handled either by power series, L'Hopital or related techniques.

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I'm interested in this path, but it takes to $\lim\limits_{x \to \infty} 5(200 \ln(z) - \ln(x) )$ – bobobobo Feb 9 '11 at 22:56
@bobobobo: I've added further details - see above. – Gone Feb 9 '11 at 23:11
Oh man!  What a good answer.  Now I have 1000 derivatives of $\lim\limits_{z \to \infty} \frac{z^{1000}}{e^{5z}}$ via L'Hos which gives $\lim\limits_{z \to \infty} \frac{1000! z^0}{1000 \times 5 e^{5z} } = 0$ – bobobobo Feb 9 '11 at 23:18
Of course, professional limit-takers know when "changes of variables" in limits is a valid step to perform, but it's surprisingly subtle to formulate it in a way that encompasses all the variants we do without thinking (and that also is still true!). Students are generally not taught changes of variables in limits very early on, if I'm not mistaken. – Greg Martin Oct 28 '11 at 20:33

Be careful using limit operation.

First, let show that $\lim_{x \rightarrow +\infty} \dfrac{\ln x}{x} = 0$. For $t \geq 1$, we have $t \geq \sqrt{t}$ which imply for $x \geq 1$

$$0 \leq \ln x = \int_1^x \dfrac{dt}{t} \leq \int_1^x \dfrac{dt}{\sqrt{t}} = 2 \sqrt{x} - 2 \leq 2 \sqrt{x}$$.

Then, for any $a,b > 0$ and $x > 1$, we have $$\dfrac{\ln^b x}{x^a} = \left( \dfrac{\ln x}{x^{\frac{a}{b}}} \right)^b = \left( \dfrac{b}{a} \right)^b \left( \dfrac{\ln (x^\frac{a}{b})}{x^{\frac{a}{b}}} \right)^b$$

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+1: I have to say, I like the second part very much, since it completely avoids L'Hopital's Rule. I'm not so happy with the first part (most people would encounter the limit of $\ln x/x$ well before encountering integrals), but the second part makes up for it. – Arturo Magidin Feb 10 '11 at 5:18

Taking $\ln^{1000}x=(\ln x)^{1000}$ you can apply L'Hopital's rule 999 times to reduce to

$$\lim_{x\to\infty}\frac{1000!\ln x}{5^{1000}x^5}.$$

Then one more application gives

$$\lim_{x\to \infty}\frac{1000!}{5^{1001}x^5}=0.$$

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You used L'Hopitals' rule once too many times. Each application reduces the exponent of $\ln(x)$ by $1$, so applying it $1000$ times reduces it to $(\ln x)^{1000-1000} = (\ln x)^0$. – Arturo Magidin Feb 9 '11 at 22:40
Thanks.  I fixed it. – Joe Johnson 126 Feb 9 '11 at 22:42
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