Finding the limit of this function as n tends to infinity... $$\lim_{n\rightarrow\infty}\frac{n}{3}\left[\ln\left(e-\frac{3}{n}\right)t-1\right]$$
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.

Your limit is equal to
$$\lim_{n\to +\infty}\frac{n}{3}\,\log\left(1-\frac{3}{en}\right),$$
but since:
$$\lim_{x\to 0}\frac{\log(1-x)}{x}=-1,$$
(by squeezing, by convexity or by De l'Hopital rule) you have:
$$\lim_{n\to +\infty}\frac{n}{3}\,\log\left(1-\frac{3}{en}\right)=-\frac{1}{e}.$$
