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How can one show that $\displaystyle{\lim_{n \to \infty}\left(2 + {1 \over n}\right)^n}$ ?.

I am trying to shoehorn the definition of $e$ somewhere but fruitlessly so.

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    $\begingroup$ $$2^n \leqslant \left(2+\frac1n\right)^n \leqslant 3^n$$ $\endgroup$ Jan 4, 2014 at 19:54
  • $\begingroup$ $\large\verb*\displaystyle*$ is not allowed in the main title. $\endgroup$ Jan 4, 2014 at 20:10
  • $\begingroup$ @user119114 : how can one show that what? What are you trying to show? $\endgroup$ Jan 5, 2014 at 4:52

4 Answers 4

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

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$$\left(2+\frac{1}{n}\right)^n = 2^n \left(1 + \frac{1/2}{n}\right)^n \sim 2^n \sqrt{e}$$

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    $\begingroup$ Awesome, thanks! $\endgroup$
    – user119114
    Jan 4, 2014 at 20:16
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    $\begingroup$ You're welcome! $\endgroup$
    – TMM
    Jan 4, 2014 at 20:17
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Since $\frac{1}{n}$ will asymptotically approach 0, the limit inside the parenthesis will go to 2. In turn, the limit of $2^n$ as $n$ approaches infinity is infinity as that function increases without bound.

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    $\begingroup$ Be very careful here with taking limits piecemeal. It doesn't cost you here, but it usually does. $\endgroup$ Jan 4, 2014 at 20:17
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Applying L'Hopital $\lim(2+\frac{1}{n})^n=\lim(e^{\log(2+\frac{1}{n})n})=e^{\lim(\log(2+\frac{1}{n})/n^{-1})}$ the exponent will blow up.

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