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This question already has an answer here:

find this limit

$$\lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}$$

my idea: $$\lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}=\lim \limits_{x\to+\infty}e^{-x}\cdot e^x=1$$

But book is answer is not 1? and How about find it? Thank you

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marked as duplicate by Guy Fsone, Alex Wertheim, Paramanand Singh limits Nov 10 '17 at 21:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The Maple command $$limit(exp(-x)*(1+1/x)^{x^2}, x = infinity) $$ produces $exp(-1/2) $ and $$asympt(exp(-x)*(1+1/x)^{x^2}, x, 3) $$ gives $$ e^{-1/2}+1/3\,{\frac {e^{-1/2}}{x}}+O \left( x^{-2} \right) .$$ $\endgroup$ – user64494 Sep 14 '13 at 11:02
  • $\begingroup$ math.stackexchange.com/questions/295584/… $\endgroup$ – Guy Fsone Nov 10 '17 at 16:26
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Since $\log(1+u)=u-\frac12u^2+o(u^2)$ when $u\to0$, $$ x^2\log\left(1+\frac1x\right)-x=x^2\left(\frac1x-\frac12\frac1{x^2}+o\left(\frac1{x^2}\right)\right)-x=-\frac12+o(1), $$ hence the limit you are after is $\mathrm e^{-1/2}$.


Note that your solution would be valid if one had $$ \lim_{x\to\infty}\frac{\left(1+\frac1x\right)^{x^2}}{\mathrm e^x}=1, $$ while in fact, $$ \lim_{x\to\infty}\frac{\left(1+\frac1x\right)^{x^2}}{\mathrm e^{x-1/2}}=1. $$


Still another take on the subject, using $O$ and $o$ Landau notations: a priori, your approach yields the estimate $$ \left(1+\frac1x\right)^{x^2}=\mathrm e^{x+o(x)}, $$ which is not even enough to show the ratio you are interested in is bounded. Strengthening a little bit the estimate, one might wish to use $$ \left(1+\frac1x\right)^{x^2}=\mathrm e^{x+O(1)}, $$ which does imply that the ratio you are interested in is bounded but not that it has a limit. What we did above is to show that $$ \left(1+\frac1x\right)^{x^2}=\mathrm e^{x-1/2+o(1)}, $$ which does imply that the ratio you are interested in is bounded, and that it has a limit, and which identifies the limit.

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  • $\begingroup$ Oh,I kown my wrong,becasuse $e^{x}\longrightarrow +\infty$? $\endgroup$ – china math Sep 14 '13 at 10:24
  • $\begingroup$ See expanded answer. $\endgroup$ – Did Sep 14 '13 at 10:31
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You guess $$ \lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}=\lim \limits_{x\to+\infty}e^{-x}\cdot e^x=1 $$ But you should add few more steps to see the mistake. You could try $$ \lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}= \left(\lim \limits_{x\to+\infty}e^{-x} \right)\left(\lim \limits_{x\to+\infty}\left(1+\dfrac{1}{x}\right)^{x^2}\right) $$ but this has indeterminate form $0 \cdot \infty$, so you cannot just multiply these two factors.

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  • $\begingroup$ +1. This way of looking at it lets you do the problem without series, using only L'Hopital's rule. In principle you could even do it without the usual "the limit is the exponential of the limit of the logarithm" argument, though the derivatives are pretty terrible if you try to do it that way. $\endgroup$ – Ian Jun 20 '14 at 15:49

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