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How is

$$\lim_{n \to \infty}2\left(1-\frac {1}{n+1}\right)^n$$ equal to the following limit

$$\lim_{n \to \infty}2e^{n\ln\left(1-\frac {1}{n+1}\right)}$$

I feel like I'm missing something. I don't get how is it possible to apply an exponential of a logarithm to the expression. Any hints?

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  • $\begingroup$ Are you missing an exponent in the first expression? $\endgroup$
    – iamwhoiam
    Nov 3, 2017 at 10:24
  • $\begingroup$ Yes @iamwhoiam, my bad! Edited. $\endgroup$
    – Cesare
    Nov 3, 2017 at 10:25
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    $\begingroup$ Note that $e^{ln(x)}=x$ $\endgroup$ Nov 3, 2017 at 10:29

4 Answers 4

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By definition we have $e^x$ and $\ln x$ are inverse functions of each other, this means that $$e^{\ln x}=x$$ and $$\ln e^x=x.$$

This is because you are 'undoing' something which you have 'done'.

So, using this, we have $$\Big(1-\frac{1}{n+1}\Big)^n=e^{\ln(1-\frac{1}{n+1})^n}=e^{n\ln(1-\frac{1}{n+1})},$$

where that last bit we also used the property of $\ln a^n=n\ln a$.

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

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The limit is $2e^{-1}$, because by the l'Hospital rule $n\ln (1-\frac{1}{n+1}) \to -1$.

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$y_n:= 2(1-1/(n+1))^n = $

$2\exp(\log(1-1/(n+1))^n=$

$2\exp(n\log(1-1/(n+1)).$

Find $\lim_{n \rightarrow \infty} n\log(1-1/(n+1)) $,

and use the continuity of the exponential

function to get $\lim_{n \rightarrow \infty} y_n.$

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