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How would you prove the following? $\lim_{n \rightarrow \infty} (1+c/n^2 - 2/n)^n = e^{-2}$

where $c\geq 1$ is a constant.

I can find proofs online of $\lim_{n \rightarrow \infty} (1-2/n)^n = e^{-2}$ but how do you make the above rigorous?

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    Edited question slightly (but thanks).2012-12-13

2 Answers 2

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Note that

$\frac 1 {n^2}-\frac 2 n +1=\left(\frac 1 n -1\right)^2$

ADD

Let $x_n= \left(1- \frac 2 n+\frac c {n^2}\right)^n $

so that $y_n=\log x_n=n \log \left(1- \frac 2 n+\frac c {n^2}\right)$

Then

$n{\log \left(1- \frac 2 n+\frac c {n^2}\right)}=$

$n\left(\frac c {n^2}- \frac 2 n\right)\frac{\log \left(1+\left(\frac c {n^2}- \frac 2 n\right)\right)}{\left(\frac c {n^2}- \frac 2 n\right)}=$

$\left(\frac c {n}- 2\right)\frac{\log \left(1+a_n\right)}{a_n}=$

Now $a_n\to 0$; so that

$\left(\frac c {n}- 2\right)\frac{\log \left(1+a_n\right)}{a_n}\to (0-2)\cdot 1 =-2$

This means $x_n\to e^{-2}$

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    I am using that. You can find various proofs of that, or give one yourself.2012-12-13
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$\lim _{n\to\infty} \left(\left(1+\frac{c-2n}{n^2}\right)^{\frac{n^2}{c-2n}}\right)^{\frac{c-2n}{n}}$

$= \left(\lim _{n\to\infty}\left(1+\frac{c-2n}{n^2}\right)^{\frac{n^2}{c-2n}}\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$

$= \left(\lim _{m\to\infty}\left(1+\frac1m\right)^m\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$ where $m=\frac{n^2}{c-2n},n\to \infty\implies m\to\infty$

$=e^{-2}$ for any finite value of $c$

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    @Sprong, please use logarithm if you are not comfortable with exponent.2012-12-13