Question

The limit as b approaches infinity of ln(b/b+1)

Answer 1

It's part of a larger integration I'm doing... Wolfram did this thing\[\lim_{b\to \infty}\ln \left( \frac{b}{b+1} \right)=\ln \left( \lim_{b\to \infty}\frac{b}{b+1} \right)\] but I'm not sure how this is justified.

Answer 2

It's easy with L'Hopital's rule from here, but I'm just not sure about how you can move the log outside in this step.

Answer 3

That step works because the natural log function \(\ln x\) is continuous for \(x>0\).

In general, if a function is continuous over a given interval, then you can reverse the operators: \(\lim f(x)=f(\lim x)\).

Answer 4

That works with any continuous function?

Answer 5

ah, it seems I need to review my limit laws; found it. Thanks!

Answer 6

Yes, for example
\[\lim_{x\to\infty}\cos\left(\frac{1}{x}\right)=\cos\left(\lim_{x\to\infty}\frac{1}{x}\right)\]
since cosine is continuous for all \(x\). If this limit was approaching 0, you could still apply this property, but the limit would not exist because \(\dfrac{1}{x}\) is not continuous at 0.

Answer 7

so append "as long as it exists" to the law above; makes perfect sense, thanks for your help.

Answer 8

You're welcome!