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$$\lim_{n\rightarrow\infty}\frac{n}{3}\left[\ln\left(e-\frac{3}{n}\right)t-1\right]$$

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.

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I don't know why you differentiated the function. If you are going to use L'Hospital's Rule, write the function as $\frac{\ln(e-3/n)-1}{3/n}$ and let $x=1/n$, or, better, $3/n$. Then want $\lim_{x\to 0^+}\frac{\ln(e-x)-1}{x}$. Now straight L'Hospital. –  André Nicolas Oct 25 '12 at 10:03
    
@AndréNicolas That's just how lost I was...*embarrassed* thank you, though! –  Siyanda Oct 25 '12 at 17:15

2 Answers 2

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?

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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: $$\lim_{n\to +\infty}\frac{n}{3}\,\log\left(1-\frac{3}{en}\right)=-\frac{1}{e}.$$

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