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Evaluate: $\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}$

attemp: Take $P=\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}$ . Then taking log both side .$\ln P=-\lim_{n \to \infty}n\ln [(1+\frac{1}{n})^n-(1+\frac{1}{n})]$.Then stuck. Please help.

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    yes.It is equal to **e**.2012-12-21

2 Answers 2

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$(1+\frac{1}{n})^n$ tends to $e$ as $n$ increases, so for big enough $n$ it's between $2$ and $3$. Then you can continue with squeeze theorem for example. We have $\frac{3}{2} \le 2 - (1+\frac{1}{n})\le(1+\frac{1}{n})^n-(1+\frac{1}{n}) \le 3 -(1+\frac{1}{n})\le 2$ for big $n$. We see that $\lim_{n \to \infty}(\frac{3}{2})^{-n} = \lim_{n \to \infty}(2)^{-n} = 0$, so $\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}$ is also $0$

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    Please explain it more.2012-12-21
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What you have done is wrong because you assumed the limit existed.

We must evaluate $\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}= \lim_{n \to \infty}e^{\log[(1+\frac{1}{n})^n-(1+\frac{1}{n})](-n)} $ Remember $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$ Thus, because $e>2$ $\lim_{n \to \infty}\log[(1+\frac{1}{n})^n-(1+\frac{1}{n})](-n)=\log(e-1)(-\infty)=-\infty $ By continuity of $e^x$, $\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}= \lim_{n \to \infty}e^{\log[(1+\frac{1}{n})^n-(1+\frac{1}{n})](-n)}=e^{-\infty}=0 $

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    @DonAntonio Oh my! That's a serious mistake. I will correct this immediately2012-12-21