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How one can evaluate following limit: $$\lim_{x\to\infty} x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)?$$

I've found this exercise in the chapter about Taylor series, but I have no idea how to solve it.

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Let $x=1/h$ i.e as $x\rightarrow \infty$, $1/h\rightarrow 0$. The limit is then

$$\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-\left({1+h}\right)^{-1/h}\right)=\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-e^{-\frac{1}{h}\ln\left({1+h}\right)}\right)$$ Also, $$\ln(1+h)=h-\frac{h^2}{2}+\frac{h^3}{3}-\cdots \Rightarrow -\frac{1}{h}\ln(1+h)=-1+\frac{h}{2}-\frac{h^2}{3}+\cdots$$ $$\Rightarrow e^{-\frac{1}{h}\ln(1+h)}=e^{-1}e^{\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)}=\frac{1}{e}\left(1+\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)+\frac{\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)^2}{2!}+\cdots\right)$$ Collecting only the terms linear in $h$, $$e^{-\frac{1}{h}\ln(1+h)}=\frac{1}{e}\left(1+\frac{h}{2}+\cdots \right)$$ Plugging this in the expression we have to evaluate, $$\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-e^{-\frac{1}{h}\ln\left({1+h}\right)}\right)=\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-\frac{1}{e}\left(1+\frac{h}{2}+\cdots\right)\right)$$ $$=\lim_{h\rightarrow 0}\frac{-1}{2e}+(\text{terms containing h})=\boxed{\dfrac{-1}{2e}}$$

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Using the Taylor series for $\log(1+x)$ and $e^x$, $$ \begin{align} \left(\frac{x}{x+1}\right)^{\large x} &=\left(1+\frac1x\right)^{\large-x}\\[6pt] &=e^{-x\log(1+\frac1x)}\\[6pt] &=e^{-x\left(\large\frac1x-\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)}\\[9pt] &=e^{-1}\,e^{\large\frac1{2x}+O\left(\frac1{x^2}\right)}\\[9pt] &=\frac1e\left(1+\frac1{2x}+O\left(\frac1{x^2}\right)\right)\tag{1} \end{align} $$ Applying $(1)$, we get $$ \begin{align} \lim_{x\to\infty}x\left(\frac1e-\left(\frac{x}{x+1}\right)^{\large x}\right) &=\lim_{x\to\infty}\left(-\frac1{2e}+O\left(\frac1x\right)\right)\\ &=-\frac1{2e}\tag{2} \end{align} $$

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  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn May 7 '14 at 23:15
  • $\begingroup$ How did you make the transition to (1) from the former line? You just dropped the $e$.. $\endgroup$ – Stabilo Jun 23 '15 at 20:27
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    $\begingroup$ @user3697301: $e^{-1}\mapsto\frac1e$ and $e^{\frac1{2x}+O\left(\frac1{x^2}\right)}\mapsto1 +\frac1{2x}+O\left(\frac1{x^2}\right)$ $\endgroup$ – robjohn Jun 23 '15 at 21:05
  • $\begingroup$ But why can you replace the second one? Thanks $\endgroup$ – Stabilo Jun 23 '15 at 21:21
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    $\begingroup$ Because $e^x=1+x+O\left(x^2\right)$ $\endgroup$ – robjohn Jun 23 '15 at 21:25
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Write $$ \frac1e-\left(\frac{x}{x+1}\right)^x=\left(e^{\frac1x}\right)^{-x}-\left(1+\frac{1}{x}\right)^{-x} $$ By the mean value theorem, $$ a^{-x}-b^{-x}=(-x)c^{-x-1}(a-b) $$ with some $c$ between $a$ and $b$. Now $$ a(x)-b(x)=e^{\frac1x}-1-\frac1x=\frac1{2x^2}+\frac1{6x^3}+..., $$ $c(x)\to1$ and $c(x)^x\to e$ by the squeeze lemma and thus $$ x\left(\frac1e-\left(\frac{x}{x+1}\right)^x\right) =-\frac1{e+o(\frac1x)}\cdot \left(\frac12+\frac1{6x}+O\left(\frac1{x^2}\right)\right)\to-\frac1{2e} $$

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  • $\begingroup$ The Mean Value Theorem works well here. Nice approach (+1) $\endgroup$ – robjohn May 6 '14 at 23:36

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