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$$\lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}$$

Can you tell me how to solve this without L'Hôpital's rule? Subsituting $+∞$ I get $\frac{∞}{∞}$ in the exponent.

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closed as off-topic by Xander Henderson, Simply Beautiful Art, RRL, GNUSupporter 8964民主女神 地下教會, José Carlos Santos Apr 29 '18 at 19:04

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  • 3
    $\begingroup$ If $x\to\infty$, $(x-1)/(x+1)\to1$. $\endgroup$ – Lord Shark the Unknown Apr 29 '18 at 15:31
  • $\begingroup$ You can divide the numerator and denominator of the exponent by $x$. $\endgroup$ – Aurel Apr 29 '18 at 15:33
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Observe that $$ xe^{\frac{x-1}{x+1}}>x,\quad x>1, $$ giving $$ \lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}\ge \lim_{x\to\ + ∞}x. $$

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$$ \ln(xe^{\frac{x-1}{x+1}})=\ln x+\frac{x-1}{x+1}\rightarrow\infty\quad\text{as} \quad x\rightarrow\infty $$

$$ \lim\limits_{x\rightarrow\infty}xe^{\frac{x-1}{x+1}}=e^{\lim\limits_{x\rightarrow\infty}\ln(xe^{\frac{x-1}{x+1}})}\rightarrow\infty $$

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$$\lim_{x\to\ + ∞} {\frac{x-1}{x+1}}= 1$$ Thus $$\lim_{x\to\ + ∞} xe^{(x-1)/(x+1)}=\lim_{x\to\ + ∞} xe=+\infty$$

Or: $$\lim_{x\to\ + ∞} xe^{\frac{x-1}{x+1}}=\lim_{x\to\ + ∞} e^{\log(x\exp((x-1)/(x+1)))}=\lim_{x\to\ + ∞} e^{(x-1)/(x+1)+\log x}=e^{1+\infty}=+\infty$$

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