# You know it’s a bad sign…

You know you’re a bit rusty from the summer when a teacher asks you to — without using L’Hopital — to prove that: $\lim_{x \rightarrow \infty} \frac{\ln(2x)}{\ln(3x)}=1$

And you are like, oh, that’s easy.

And then — after two false starts — it takes you 3 minutes to figure out.

Just remember: you are a calculus teacher.

I’m going to say it again: you are a calculus teacher.

Maybe if I say it enough times, it will be true.

The technique for this problem is to substitute $u=1/x$. Then a short bit of manipulation later, you’re there.

1. Adam Glesser says:

How about if you break it up as:
ln(2)/ln(3x) + ln(x)/[ln(3) + ln(x)] =
= ln(2)/ln(3x) + 1/[(ln(3)/ln(x)) + 1]
As x->infinity, the first goes term goes to 0, the second term to 1/1 = 1.

2. samjshah says:

Ha! I did exactly those steps. Except I first made that substitution. Which was totally lame. Because you didn’t need that substitution at all.

Well, I guess there’s more than one way to skin a cat.

Your way: put the cat to sleep. Skin cat.

My way: put the cat to sleep. Put the cat to sleep again. Skin cat.

3. mbork says:

And I would do it in a yet different way: [ln(2)+ln(x)]/[ln(3)+ln(x)], and then: [ln(2)/ln(x)+1]/[ln(3)/ln(x)+1]. The first term in both the numerator and the denominator tends to zero, and the second ones are constant – voila!

In fact, this is the way we usually do this in schools in Poland. The method is: find the “greatest” (in an asymptotic sense) term in both numerator and the denominator, and if it’s the same – divide both of them by that term; if it’s not the same, usually you get 0 or infinity. The method works fine with rational expressions and also in some other cases, and is easy to remember.

4. samjshah says:

Oh, that’s nice! It works with rational functions, so why not others? I’m using this when we get to limits.