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How can I compute this limit of the sequence? $$\lim_{n\to\infty}{(\sqrt[n]{e}-\frac{2}{n})^n}$$

Breldor
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3 Answers3

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This depends on the tools you master, but if you are aware that $$\sqrt[n]{\mathrm e}=\mathrm e^{1/n}=1+1/n+o(1/n),$$ then you can write that $$(\sqrt[n]{\mathrm e}-2/n)^n=(1-1/n+o(1/n))^n\to \mathrm e^{-1}.$$

Did
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  • Is there some obvious justification for the last step? (other than using $(1 - 1/n + f(n))^n = ((1-1/h(n))^{h(n)})^{1 -nf(n)} \to e^{-1}$ if $nf(n) \to 0$ (as $h(n) \to \infty$)? (or taking logs...) – Aryabhata Jan 07 '15 at 02:38
  • @Aryabhata Well, you gave two of them... – Did Jan 07 '15 at 07:19
  • The reason I ask is this answer in the dupe question: http://math.stackexchange.com/a/1088943/1102. I was thinking perhaps I was missing some nice theorem or something... – Aryabhata Jan 07 '15 at 19:32
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With the substitution $x=1/t$ we have $$ \lim_{x\to\infty}(e^{1/x}-2/x)^x= \lim_{t\to0^+}(e^t-2t)^{1/t} $$ The limit of the logarithm of the expression $(e^t-2t)^{1/t}$ is $$ \lim_{t\to0^+}\frac{\log(e^t-2t)}{t} \overset{\mathrm{(H)}}{=} \lim_{t\to0^+}\frac{e^t-2}{e^t-2t}=-1 $$ Without l'Hôpital, recall that $\log(e^t-2t)=\log(1-t+o(t))=-t+o(t)$.

egreg
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Another way you can go is: $$\begin{align} \lim_{n\to\infty}\log\left(\sqrt[n]e-\frac2n\right)^n & =\lim_{n\to\infty}{\log\left(e^{1/n}-\frac2n\right)\over\frac1n}\\ & = \lim_{n\to\infty}{\left(-\frac{1}{n^2}e^{1/n}+\frac{2}{n^2}\right)/\left(e^{1/n}-\frac{2}{n}\right)\over -\frac{1}{n^2}} \end{align}$$,

by L'Hospital's Rule. Now multiply numerator and denominator by $-n^2$, to get $${e^{1/n}-2\over e^{1/n}-\frac2n}$$ and it should be fairly easy.

Tim Raczkowski
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