Question

Can I do this limit without doing the \( t=\frac1n\) thing?\[\lim_{n\to\infty}n\ln{n\over n+1}\]

Answer 1

Any particular reason for which you don't want to use the obvious substitution?

Answer 2

because it leads to l'Hospital, which is wolfram's method of choice (and a bit messy in this case)
usually wolf does things the long way, so I was asking if there is a smarter way
if not, so be it...

Answer 3

\[\lim_{n\rightarrow\infty}n\ln{\frac{n}{n+1}}=\lim_{n\rightarrow\infty}n\ln{\left(1-\frac{1}{n+1}\right)}=\lim_{n\rightarrow\infty}n\left(-\frac{1}{n+1}\right)=-1\]