$$\lim_{x\rightarrow\pm\infty}\left(e^{x/(x+1)}(1-x)+e\cdot x\right)$$
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.
$\left(x\left(e -e^{\dfrac{x}{x+1}}\right)\right) = e\left(x\left(1 -e^{-\dfrac{1}{x+1}}\right)\right) = e \dfrac{e^{-\dfrac{1}{x+1}}-1}{-\dfrac{1}{x+1}}\dfrac{x}{x+1} \to e (e^x)'|_{x=0}= e$