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I am trying to evaluate this limit:

$$\lim_{n\to\infty}\left[e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\right].$$

I saw it is a limit of $1^\infty$ type and I tried to evaluate it like this: $(1-\frac{1}{\sqrt{n}})^n=e^\frac{-n}{\sqrt{n}}=e^{-\sqrt{n}}$ and that means the limit equals $e^0=1$. But the answer in my textbook is $\frac{1}{\sqrt{e}}$. Am I not allowed to split the limit like that?

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  • $\begingroup$ limit for $n\to \infty$? $\endgroup$ – Riccardo.Alestra Jul 6 '17 at 10:22
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    $\begingroup$ No you are not allowed to split it like that. Also $(1-1/\sqrt n)^n=e^{-n/\sqrt n}$ is very wrong to begin with. I guess you want to write $\to$ instead of $=$ but still you are not allowed to split here. $\endgroup$ – M. Winter Jul 6 '17 at 10:24
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    $\begingroup$ What you are really doing when splitting a limit is making the simplification $$\lim_{n\to\infty}\left(e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\right) \overset{?}{=} \left(\lim_{n\to\infty}e^{\sqrt{n}}\right) \left(\lim_{n\to\infty}\left(1-\frac{1}{\sqrt{n}}\right)^n\right),$$ which would be valid if both limits were finite. Unfortunately, the two limits are $\infty$ and $0$ respectively, so you still have an indeterminate form $\infty\cdot0$. $\endgroup$ – Rahul Jul 6 '17 at 10:34
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    $\begingroup$ @M.Winter What is your "Hint" supposed to hint at, exactly? $\endgroup$ – Did Jul 6 '17 at 10:48
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    $\begingroup$ "Am I not allowed to split the limit like that?" No you are not, basically because the splitting you suggest leads to $\infty\cdot0$, which is a well-known under-determined form (as @Rahul already explained). // To solve the limit in your question, consider the logarithm of the sequence and use the two-terms expansion $$\log\left(1-\frac1{\sqrt{n}}\right)=-\frac1{\sqrt{n}}-\frac1{2n}+o\left(\frac1n\right)$$ But be warned that this yields the answer awfully easily and quickly, so if you prefer long and tortuous roads, you might want to stay with L'Hopital or some other similar monstrosity... $\endgroup$ – Did Jul 6 '17 at 10:55
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It's simpler to find he limit of the log first, using Taylor's formula at order $2$: \begin{align} \log\Biggl(e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\Biggr)&=\sqrt n+n\log\biggl(1-\frac1{\sqrt n}\biggr)=\sqrt n-n\biggl(\frac1{\sqrt n}+\frac1{2n}+o\Bigl(\frac1{ n}\Bigr)\biggr)\\&=-\frac12+o(1). \end{align} So the log tends to $-\frac12$ when $n$ tends to $\infty$, and we conclude that $$\lim_{n\to\infty}\log\Biggl(e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n\Biggr)=\frac1{\sqrt{\mathrm e}}.$$

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  • $\begingroup$ You should have read my comment more carefully: the order 2 expansion of the logarithm is needed here... $\endgroup$ – Did Jul 6 '17 at 11:29
  • $\begingroup$ And, again, an instant unwarranted upvote... What is going on on this page? $\endgroup$ – Did Jul 6 '17 at 11:30
  • $\begingroup$ @Did: Oh! yes. Sorry, I really was out of my mind :-( I'vefixed the answer. $\endgroup$ – Bernard Jul 6 '17 at 11:32
  • $\begingroup$ @haqnatural: It's not my day! I'll fix that at once $\endgroup$ – Bernard Jul 6 '17 at 11:39
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$$\lim_{n\rightarrow+\infty}e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^n=\lim_{n\rightarrow+\infty}e^{\sqrt{n}}\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}\cdot\frac{n}{-\sqrt{n}}}=$$ $$=\lim_{n\rightarrow+\infty}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}\right)^{\sqrt{n}}=$$ $$=\lim_{n\rightarrow+\infty}\left(1+\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)^{\frac{1}{\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}\cdot\sqrt{n}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}=$$ $$=e^{\lim\limits_{n\rightarrow+\infty}\sqrt{n}\left(\frac{e}{\left(1-\frac{1}{\sqrt{n}}\right)^{-\sqrt{n}}}-1\right)}=e^{-\frac{1}{e}\lim\limits_{x\rightarrow0}\frac{e-(1+x)^{\frac{1}{x}}}{x}}$$ and the rest it is the L'Hôpital's rule.

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    $\begingroup$ This would be rather terrible if one had to go through such loops (with no clear end in view, if I am not mistaken) to solve the question, don't you think? Why not rather teach the OP the correct tools, which lead safely and directly to the answer? $\endgroup$ – Did Jul 6 '17 at 10:51
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    $\begingroup$ Indeed the approach you chose is terrible (and one can only marvel at the upvotes it collects)... All the more reason to switch to more sensible tools. $\endgroup$ – Did Jul 6 '17 at 11:07
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    $\begingroup$ the jump to the fourth line seems not justified. How you knows that the limit of the third line can be simplified to an exponential limit? $\endgroup$ – Masacroso Jul 6 '17 at 11:26
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    $\begingroup$ Hmmm... yet another trouble: how are we supposed to use L'H to identify $$\lim_{x\to0}\frac{e-(1+x)^{1/x}}{x}\ ?$$ By differentiating the numerator (and the denominator), I presume? But the whole ratio is already the opposite of the derivative of $$f(x)=(1+x)^{1/x}$$ at $0$, by definition of the derivative, so, if one is able to compute $f'(0)$, why invoke L'H at all? $\endgroup$ – Did Jul 6 '17 at 11:37
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    $\begingroup$ Such a cumbersome solution, this is hardly useful IMO. If one wanted to use L'Hopsital's, one should've taken the log of the limit and turned the result into a fraction. But by then, as @Did mentions, L'H is not necessary, as the resulting limit is of the form of $\frac{f(x)-f(h)}{x-h}$ (after some algebraic manipulation), so L'H is not needed. $\endgroup$ – Simply Beautiful Art Jul 6 '17 at 12:09
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We have $n\ln(1-\frac{1}{\sqrt{n}})-\sqrt{n}=-\frac{1}{2}+\mathcal{O}(\frac{1}{\sqrt{n}})$, so the limit is $e^{-1/2}$ as required.

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