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$\begingroup$

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

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.2012-10-25
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    @AndréNicolas That's just how lost I was...*embarrassed* thank you, though!2012-10-25

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