Find the limit of $\lim_{n \to \infty}({n\over n+1})^n$.
$$\lim_{n \to \infty}({n\over n+1})^n=\lim_{n \to \infty}({n+1-1\over n+1})^n=\lim_{n \to \infty}(1-{1\over n+1})^n$$
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Can you use L'Hospital's Rule? – graydad Jun 26 '15 at 13:40
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What is your definition of $e$ or $e^{-1}$? – Henry Jun 26 '15 at 13:41
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Do you know the definition of $e$ as a limit ? – Jun 26 '15 at 13:41
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2This is the definition of $e^{-1}$ you can't "evaluate" any more than that. – Zach466920 Jun 26 '15 at 13:41
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2@graydad how can I use L'Hospital here? – gbox Jun 26 '15 at 13:46
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4See also: http://math.stackexchange.com/questions/1350338/how-to-prove-that-lim-limits-n-to-infty-left-1-frac1n-rightn-fr and http://math.stackexchange.com/questions/596771/finding-the-limit-of-left-fracnn1-rightn – Martin Sleziak Oct 26 '15 at 06:45
3 Answers
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$$ ({n\over n+1})^n={1 \over ({n+1\over n})^n} = {1 \over (1+{1\over n})^n} \to {1 \over e} $$
lhf
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$$\begin{align}\lim_{n \to \infty}({n\over n+1})^n=\lim_{n \to \infty}({n+1-1\over n+1})^n&=\lim_{n \to \infty}(1-{1\over n+1})^n\\&=\underbrace{\lim_{n\to\infty}\left[\left(1-\frac{1}{n+1}\right)^{-(n+1)}\right]^{-\frac{n}{n+1}}}_{e^{-\frac{n}{n+1}}}\\&=\frac{1}{e}\end{align}$$
Lucas
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As for $e^{-\frac{n}{n+1}}$ we say that $lim_{n \to \infty} -{n \over n+1}=n$? – gbox Jun 26 '15 at 14:06
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2@gbox $$\lim_{n\to\infty}-\frac{n}{n+1}=-\lim_{n\to\infty}\frac{n+1-1}{n+1}=-\lim_{n\to\infty} (1-\frac{1}{n+1})=-1$$ – Lucas Jun 26 '15 at 14:07
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Observing that the exponential and logarithm are inverses of each other, you can apply L'Hopital's rule to get: \begin{equation*} \lim_{n\to\infty}(\frac{n}{n+1})^n=\exp \lim_{n\to\infty}(n\ln(\frac{n}{n+1})) =\exp (\lim_{n\to\infty}\frac{\ln(\frac{n}{n+1})}{\frac{1}{n}}) \\ = \exp(\lim_{n\to\infty} -\frac{n}{n+1}) =\exp(\lim_{n\to\infty}\frac{-1}{1+\frac{1}{n}})=\frac{1}{e}. \end{equation*}