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

Answered by Mathspace last updated on 17/Jan/22

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

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