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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5Are you sure it shouldn't be $x((1+\frac1x)^x-e)$? – hmakholm left over Monica Feb 06 '17 at 14:01
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1^ As stated, it surely diverges. – Alex Reinking Feb 06 '17 at 14:02
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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... – Did Feb 06 '17 at 14:05
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You are right, I fixed it. – user401516 Feb 06 '17 at 14:07
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The answer, by the way, should be 0. – user401516 Feb 06 '17 at 14:07
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"The answer, by the way, should be 0" No. – Did Feb 06 '17 at 14:13