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I'm trying to find the limit of: $$\lim\limits_{x \to \infty}x\left(\left(1+\frac{1}{x}\right)^x-e \right)$$

I tried to use L'Hôpital but it lead me nowhere. Any ideas?

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    Are you sure it shouldn't be $x((1+\frac1x)^x-e)$? – 2017-02-06
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    ^ As stated, it surely diverges. – 2017-02-06
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    This was asked very recently (minus the typo signalled by Henning) and it was explained how to show without computation that the limit is $$-\frac e2$$ using only the equivalents when $u\to0$: $$e^u-1\sim u\qquad\log(1+u)-u\sim-\tfrac12u^2$$ If I remember correctly, it was also mentioned that using L'H in this case leads comparatively to a mess... – 2017-02-06
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    You are right, I fixed it. – 2017-02-06
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    The answer, by the way, should be 0. – 2017-02-06
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    "The answer, by the way, should be 0" No. – 2017-02-06

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