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$$\lim_{x\rightarrow\pm\infty}\left(e^{x/(x+1)}(1-x)+e\cdot x\right)$$

Without L'Hôpital's rule. What I've done: $$e - \lim_{x\rightarrow\pm\infty}\left(-e^{x/(x+1)}\cdot x+e\cdot x\right) = e - \lim_{x\rightarrow\pm\infty}\left(x\left(e -e^{x/(x+1)}\right)\right)$$ Even though I don't know how to continue here, the weird part is that Mathematica tells me the last limit expression is equal to $e$, which would mean the original expression becomes $0$. Although if I input the original expression into Mathematica, I get the value $2e$ which makes more sense.

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$\left(x\left(e -e^{\dfrac{x}{x+1}}\right)\right) = e\left(x\left(1 -e^{-\dfrac{1}{x+1}}\right)\right) = e \dfrac{e^{-\dfrac{1}{x+1}}-1}{-\dfrac{1}{x+1}}\dfrac{x}{x+1} \to e (e^x)'|_{x=0}= e$

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  • $\begingroup$ Thank you, there's one problem remaining though. This gets me $e-e = 0$. While I'm supposed to get the answer $2e$. But thank you, that was very helpful! $\endgroup$ – B. Lee Nov 13 '14 at 22:44
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    $\begingroup$ @XMLParsing Recheck your simplification, you should get $e+ \lim\cdots$, so the result is indeed $2e$ $\endgroup$ – Petite Etincelle Nov 13 '14 at 22:46
  • $\begingroup$ Hmm, one question: When $x$ approaches $\pm\infty$ how come $\dfrac{e^{-\dfrac{1}{x+1}}-1}{-\dfrac{1}{x+1}}$ goes to $1$ and not $\infty$? I know about the common limit $\lim_{x\rightarrow0}\dfrac{e^x-1}{x} = 1$ but that's $0$, not $\infty$ $\endgroup$ – B. Lee Nov 13 '14 at 22:56
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    $\begingroup$ @XMLParsing as you write $\dfrac{e^h-1}{h} \to 1$ when $h\to 0$, now take $h = -\dfrac{1}{x+1}$ $\endgroup$ – Petite Etincelle Nov 13 '14 at 22:58

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