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Question Number 164447 by LEKOUMA last updated on 17/Jan/22

Answered by Mathspace last updated on 17/Jan/22

f(x)=(e^(ln^2 x) /e^(xln(lnx)) )=e^(ln^2 x−xln(lnx))   u(x)=ln^2 x−xln(lnx)=_(lnx=t)   =t^2 −e^t lnt =e^t (t^2 e^(−t) −lnt)  ∼−e^t lnt→−∞ ⇒  lim_(x→+∞) f(x)=0

$${f}\left({x}\right)=\frac{{e}^{{ln}^{\mathrm{2}} {x}} }{{e}^{{xln}\left({lnx}\right)} }={e}^{{ln}^{\mathrm{2}} {x}−{xln}\left({lnx}\right)} \\ $$$${u}\left({x}\right)={ln}^{\mathrm{2}} {x}−{xln}\left({lnx}\right)=_{{lnx}={t}} \\ $$$$={t}^{\mathrm{2}} −{e}^{{t}} {lnt}\:={e}^{{t}} \left({t}^{\mathrm{2}} {e}^{−{t}} −{lnt}\right) \\ $$$$\sim−{e}^{{t}} {lnt}\rightarrow−\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{0} \\ $$

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