0
$\begingroup$

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.

  • 0
    Is $\ln^{1000}x$ intended to mean $(\ln x)^{1000}$? That's usual withtrigonometric functions, but I had never seen it used with logarithms...2011-02-09
  • 0
    The limit after "Here's what I get" it not at all the same as what you started with.2011-02-10
  • 0
    @Hans: Why? It was writing $e^{ln{y}}=y$2011-02-13
  • 0
    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 ...".2011-02-13

3 Answers 3

3

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

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

  • 0
    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 $(\ln x)^{1000-1000} = (\ln x)^0$.2011-02-09
6

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 $$ \dfrac{\ln^b x}{x^a} = \left( \dfrac{\ln x}{x^{\frac{a}{b}}} \right)^b = \left( \dfrac{b}{a} \right)^b \left( \dfrac{\ln (x^\frac{a}{b})}{x^{\frac{a}{b}}} \right)^b $$

which answer your question.

  • 0
    +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.2011-02-10
3

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.

  • 0
    I'm interested in this path, but it takes to $ \lim\limits_{x \to \infty} 5(200 \ln(z) - \ln(x) ) $2011-02-09
  • 0
    @bobobobo: I've added further details - see above.2011-02-09
  • 0
    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 $2011-02-09
  • 0
    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.2011-10-28