Why is the limit of ln(x)/arctan(x) as x approaches 0 from the right, not zero?

[Numnum]

Homework Statement

Why is the limit of ln(x)/arctan(x) as x approaches 0 from the right, not zero?

Homework Equations

The Attempt at a Solution

I used L'hopital's rule and got zero. But question specifically states that is not the answer.

 

[jbunniii]

$$\lim_{x \rightarrow 0^+} \ln(x) = -\infty$$
whereas
$$\lim_{x \rightarrow 0^+} \arctan(x) = 0$$
so L'Hopital's rule does not apply.

 

[HallsofIvy]

As jbunniii said, L'Hopital's rule does not apply here. Notice that If x= .000001, we have
\frac{ln(.000001)}{arctan(.000001)}= \frac{-13.8155}{.000001}= -13815510
not anywhere near 0!

 

[Numnum]

Thank you!

But, that's all there is to it? I don't have to simplify anything? Just plug in values?

 

[HallsofIvy]

No, I didn't say that. In order to prove that a limit is a specific number you have to prove that it gets arbitrarily close to that number. My point was that for x very close to 0, the function value is very far away from 0. It is theoretically possible that a function would turn from being around -13 million at x= .000001 to 0 at x= 0, but that would be a very strange function!

 

[haruspex]

It's so obviously unbounded that it's rather a strange question. Are you sure you've stated it correctly?