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

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.

-
Do you know what $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ is? –  Michael Albanese Dec 21 '12 at 12:33
yes.It is equal to e. –  A.D Dec 21 '12 at 12:34

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 $e^x$, $$\lim_{n \to \infty}[(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}= \lim_{n \to \infty}e^{\log[(1+\frac{1}{n})^n-(1+\frac{1}{n})](-n)}=e^{-\infty}=0$$

-
@Jabali: You wanted Adam to explain more but you have a complete answer here! :-) –  Babak S. Dec 21 '12 at 12:50
I don't understand adam's word.So I asked to explain.But I don't understand here how I conclude that limit does not exist. –  A.D Dec 21 '12 at 13:02
@DonAntonio Oh my! That's a serious mistake. I will correct this immediately –  Nameless Dec 21 '12 at 14:53

$(1+\frac{1}{n})^n$ 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 $0$

-
Please explain it more. –  A.D Dec 21 '12 at 12:41