$\lim_{n\rightarrow\infty}\frac{n}{3}\left[\ln\left(e-\frac{3}{n}\right)t-1\right]$
I'm having little trouble figuring this out. I did try to differentiate it about 3 times and ended up with something like this
$f'''(n) = \frac{1}{3} \left(\frac{1}{e - \frac{3}{n}}\right) + \left(\frac{1}{3e - \frac{9}{n}}\right) - \left(\frac{9}{3e - \frac{9}{n}}\right)$
So I wonder if the limit of this would be calculated as
$lim_{n\rightarrow\infty}\frac{1}{3} \left(\frac{1}{e - \frac{3}{n}}\right) + \left(\frac{1}{3e - \frac{9}{n}}\right) - \left(\frac{9}{3e - \frac{9}{n}}\right) = \frac{-7}{3e}$
Which feels terribly wrong.
I suspect that I did the differentiation wrong.
Any pointers would be cool and the provision of a simpler method would be dynamite.