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I tried to compute a limit: $$\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)$$ What I've done is that: $$\begin{align} &\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)\\ =&\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^{x+1}\left(\frac{x+1}{x}\right)\right)\\ =&\lim_{x\to\infty}\left(1-\frac 1{x+1}\right)^{x+1}x\left(1-\frac{x+1}{x}\right)\\ =&-\frac1e \end{align}$$ But Wolfram|Alpha seems not to agree with me, in which it gave $$-\frac1{2e}$$ So what actually does the error come from in my approach? And how should I compute the limit? Thanks in advance.

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marked as duplicate by Lord Shark the Unknown, Daniel W. Farlow, hardmath, Yujie Zha, Henrik Jun 25 '17 at 15:28

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  • $\begingroup$ @LordSharktheUnknown But I still want explanation that where did the error in my attempt come from. $\endgroup$ – BAI Jun 25 '17 at 13:51
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    $\begingroup$ An $e$ mysteriously disappears between line 2 and line 3; indeed I am at a loss as to where line 3 comes from. $\endgroup$ – Lord Shark the Unknown Jun 25 '17 at 13:54
  • $\begingroup$ @LordSharktheUnknown that is where I think I might have an error, that I put $1/e=\lim_{x\to\infty}(1-\frac{1}{x+1})^{x+1}$ $\endgroup$ – BAI Jun 25 '17 at 13:56
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We'll make use of this standard limit : $$\lim_{x\to0}\frac{e^x-1}{x}=1$$ Therefore, $$\lim_{x\to\infty}x\bigg(\frac{1}{e}-\bigg(\frac{x}{x+1}\bigg)^x\bigg)=\lim_{x\to\infty}\frac{\frac{1}{e}-\big(\frac{x}{x+1}\big)^x}{\frac{1}{x}}=\lim_{x\to\infty}\frac{e^{\ln\frac{1}{e}}-e^{x\ln\big(\frac{x}{x+1}\big)}}{\frac{1}{x}} = \lim_{x\to\infty}e^{x\ln\big(\frac{x}{x+1}\big)} \cdot\frac{e^{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}-1}{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}\cdot\frac{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}{\frac{1}{x}}$$ Which happens to be $$l = \frac{1}{e}\lim_{x\to\infty}\frac{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}{\frac{1}{x}} = \lim_{y\to0_+}\frac{1}{e}\frac{y\ln\frac{1}{e}-\ln\frac{1}{1+y}}{y^2}=\lim_{y\to0_+}\frac{1}{e}\frac{\ln(1+y)-y}{y^2}=-\frac{1}{2e}$$ The last limit I guess you could solve using the technique here : Are all limits solvable without L'Hôpital Rule or Series Expansion, but it's easier to just apply l'Hospital once and some simple fraction manipulations.

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