I could transform it to $$\lim_{x\to\infty}x\left(\dfrac{1}{e}-\left(1+\dfrac{1}{x}\right)^{-x}\right).$$
I think using sandwich theorem would help but I was unable to find the bounds.
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Are you allowed to use L'hopitals rule? – ASKASK Jun 18 '15 at 07:21
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Yes. But I would prefer a solution without using it. – Grobber Jun 18 '15 at 07:27
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Another way based on Taylor expansions : consider $$A=\left(\dfrac{x}{x+1}\right)^x$$ Taking logarithms $$\log(A)=x\log\left(\dfrac{x}{x+1}\right)=-x\log\left(\dfrac{x+1}{x}\right)=-x\log\left(1+\dfrac{1}{x}\right)$$ Now, remember that, for small $y$, $$\log(1+y)=\frac {y}{1}-\frac {y^2}{2}+\frac {y^3}{3}+\cdots$$ Replace $y$ by $\frac {1}{x}$ which makes $$\log(A)=x\left(\frac {1}{x}-\frac {1}{2x^2}+\frac {1}{3x^3}+\cdots\right)=1-\frac {1}{2x}+\frac {1}{3x^2}+\cdots$$ So $$A=e \,e^{-\frac {1}{2x}+\frac {1}{3x^2}+\cdots}$$ Now, remember that, for small $y$, $$e^y=1+\frac{y}{1}+ \frac{y^2 }{2}+\cdots$$ Replace $y$ by $-(\frac {1}{2x}-\frac {1}{3x^2})$ and you should arrive to $$A=\frac{1}{e}+\frac{1}{2 e x}-\frac{5}{24 e x^2}+\cdots$$
I am sure that you can take from here (and show not only the limit but also how it is approached).
Claude Leibovici
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