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I'm quite unfamiliar with the tricky techniques to compute limits. I had learned analysis, so I want to know the every detail to compute $\lim_{n \rightarrow\infty}\ln\left(\frac{n-1}{n+1}\right)^n$, for example what theorems involved, even though it may usually be omitted or overlooked in introductory calculus courses. Thanks.

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    $\begingroup$ Is it $\ln\left(\left(\frac{n-1}{n+1}\right)^n\right)$ or $\left(\ln\left(\frac{n-1}{n+1}\right)\right)^n$ ?? $\endgroup$ – Watson Sep 4 '16 at 13:11
  • $\begingroup$ it's the former $\endgroup$ – Eric Sep 4 '16 at 13:23
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I would normally use E.H.E.'s method to do this, but here is an alternate way, using L'Hospital's rule:

$\displaystyle\lim_{x\to\infty}\ln\left(\frac{x-1}{x+1}\right)^x=\lim_{x\to\infty}x\ln\left(\frac{x-1}{x+1}\right)=\lim_{x\to\infty}\frac{\ln(x-1)-\ln(x+1)}{1/x}=\lim_{x\to\infty}\frac{\frac{1}{x-1}-\frac{1}{x+1}}{-\frac{1}{x^2}}$

$\displaystyle=\lim_{x\to\infty}\frac{-2x^2}{x^2-1}=\lim_{x\to\infty}\frac{-2}{1-\frac{1}{x^2}}=-2$

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  • $\begingroup$ It is exactly what I waiting for!!!! How elegant is this solution! Thanks!! $\endgroup$ – Eric Sep 5 '16 at 4:07
  • $\begingroup$ @Eric You're welcome. Using the same ideas, you can show that $\ln(1+\frac{a}{n})^n\to a$, so $(1+\frac{a}{n})^n\to e^{a}$. $\endgroup$ – user84413 Sep 5 '16 at 21:10
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As written, the limit is $0$.

First, we have that $$ \frac{x}{1+x}\le\log(1+x)\le x $$ so that $$ \frac2{n+1}\le\log\left(\frac{n+1}{n-1}\right)=\log\left(1+\frac2{n-1}\right)\le\frac2{n-1} $$ Therefore, $$ \begin{align} \left|\,\lim_{n\to\infty}\left[\log\left(\frac{n-1}{n+1}\right)\right]^n\,\right| &=\lim_{n\to\infty}\left[\log\left(1+\frac2{n-1}\right)\right]^n\\ &\le\lim_{n\to\infty}\left[\frac2{n-1}\right]^n\\[4pt] &=0 \end{align} $$


However, perhaps the following was intended $$ \begin{align} \lim_{n\to\infty}\log\left[\left(\frac{n-1}{n+1}\right)^n\right] &=\lim_{n\to\infty}\log\left[\left(1-\frac2{n+1}\right)^{n+1}\left(1+\frac2{n-1}\right)\right]\\ &=\log\left(e^{-2}\cdot1\right)\\[6pt] &=-2 \end{align} $$

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  • $\begingroup$ Could you give the detail from the line 2 to line 3 in first part? $\endgroup$ – Eric Sep 4 '16 at 13:34
  • $\begingroup$ I was working on an edit to do just that. $\endgroup$ – robjohn Sep 4 '16 at 13:39
  • $\begingroup$ Is there a way avoiding Taylor expansions? $\endgroup$ – Eric Sep 4 '16 at 13:43
  • $\begingroup$ I have removed the big-O estimate. However, in a comment you indicated that you were interested in the second interpretation, not the first. $\endgroup$ – robjohn Sep 4 '16 at 16:11
  • $\begingroup$ I don't understand the second part. How could you know this way will work? I mean you change the original fraction into $(1-\cdots)^{\cdots}$, and then it will equals to some type of $e^n$. What if the original can't change to $(1-\cdots)^{\cdots}$, but $(100-\cdots)^{\cdots}$? $\endgroup$ – Eric Sep 5 '16 at 4:00
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Hint

$$\left(\frac{n-1}{n+1}\right)^n=\frac{(1-1/n)^n}{(1+1/n)^n}$$ then use $$\lim_{n\rightarrow \infty }(1\pm\frac{1}{n})^n=e^{\pm1}$$

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  • $\begingroup$ Is there a way avoiding using $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$? $\endgroup$ – Eric Sep 4 '16 at 13:45
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    $\begingroup$ @Eric You want to avoid the all powerful, elementary and very important limit $\;\lim\left(1+\frac1n\right)^n\;$ , and also to avoid Taylor expansion...what's next? To avoid using mathematics altogether? $\endgroup$ – DonAntonio Sep 4 '16 at 13:53
  • $\begingroup$ Both of them require too many preliminary results to have been first proved. Notice that some authors didn't take the limit of the sequence as the definition of $e$, especially in the analysis book. The fact that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ is also a theorem under this logical order. So it's natural to ask if there's a proof of the problem that doesn't use that limit characterization of $e$. $\endgroup$ – Eric Sep 4 '16 at 18:12
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This might be useful $${n-1\over n+1} = 1-{2\over n+1 }.$$

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    $\begingroup$ How could you come up with this? Is it a somewhat specific and daily-use technique? $\endgroup$ – Eric Sep 4 '16 at 13:47
  • $\begingroup$ You could either use Long Division or the trick $\frac{n+1-2}{n+1}$ to see this. $\endgroup$ – yoyostein Sep 4 '16 at 14:12
  • $\begingroup$ I mean what's the next step you gonna take?(What's the reason that make you want to try do so.) In general, when will we do the "Long Division" to a fraction in dealing with sequence? Is there a systematic guidelines? $\endgroup$ – Eric Sep 5 '16 at 4:03
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\begin{align} \lim_{x\to\infty}\ln\left(\frac{x-1}{x+1}\right)^x &= \lim_{x\to\infty}x\ln\frac{x-1}{x+1} \\[6px] (t=1/x)\qquad &= \lim_{t\to0^+}\frac{1}{t}\ln\frac{1-t}{1+t} \\[6px] &=\lim_{t\to0^+}\frac{\ln(1-t)-\ln(1+t)}{t} \\[6px] (\text{Taylor})\qquad &=\lim_{t\to0^+}\frac{(-t+o(t))-(t+o(t))}{t} \\[6px] &=-2 \end{align}

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Use the log limit $$\ln x=\lim_{n \to \infty}n(x^{\frac 1 n}-1)$$

Then $$\lim_{n \rightarrow\infty}\ln\left(\frac{n-1}{n+1}\right)^n=\lim_{n \to \infty}n\left(\left(\frac{n-1}{n+1}\right)^{\frac n n}-1\right )$$

$$=\lim_{n \to \infty}n\left(\left(1-\frac{2}{n+1}\right)-1\right )$$

$$=\lim_{n \to \infty}\left(\frac{-2n}{n+1}\right )$$

$$=\lim_{n \to \infty}\left(\frac{-2}{1+\frac 1n}\right )$$

$$=-2$$

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