You know you’re a bit rusty from the summer when a teacher asks you to — without using L’Hopital — to prove that:
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 
Continuous Everywhere but Differentiable Nowhere 
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.
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.
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.
Oh, that’s nice! It works with rational functions, so why not others? I’m using this when we get to limits.