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Show with help of the Bernoulli Inequality that $$\lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)^{n}=1$$ End with: $$\lim_{n\rightarrow\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}$$

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2 Answers 2

Note that Bernoulli's Inequality says that $(1+x)^n \ge 1+ nx$ if $x\ge -1$. In particular, let $x=-\dfrac{1}{n^2}$. We get $$1-\frac{n}{n^2} \le \left(1-\frac{1}{n^2}\right)^n \lt 1.$$ Now use Squeezing to conclude that the limit is $1$.

Remark: If we have defined $e$ by $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n,$$ then we can use the above result to conclude that since $$\left(1-\frac{1}{n^2}\right)^n=\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n,$$ we must have $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}.$$

This is a useful step in one approach to defining the exponential function.

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Bernoulli's ineq. must be $\,(1+x)^n\geq 1+nx\,$ . Of course, +1 anyway –  DonAntonio Oct 17 '12 at 19:06
    
@DonAntonio: Thanks. Such a short answer but still a typo! –  André Nicolas Oct 17 '12 at 19:13
    
In your last, you show $e = \frac 1e$?! –  Ross Millikan Oct 17 '12 at 22:44
    
@RossMillikan: Thanks. The perils of cut and paste! I copied the third to last displayed formula, with intent to modify. Did not modify enough. –  André Nicolas Oct 17 '12 at 23:34
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To ask the second question, if we can use that $$ \lim_{n\to\infty}\left(1+\frac{a}{n} \right)=e^a, $$ then $$ \lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)^{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{-1}{n}\right)^{n}=e\cdot e^{-1}=1. $$

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