$$\lim_{n\rightarrow\infty}\frac{n}{3}\left[\ln\left(e-\frac{3}{n}\right)t-1\right]$$
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.
Best Answer
Lets write $f(x) := \log(e - x)$, then your limits reads (as $f(0) = \log e = 1$) \[ \lim_{n \to \infty} \frac{f(\frac 3n) - f(0)}{\frac 3n} \] Do you have seen such a limit before?