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Question Number 32990 by abdo imad last updated on 09/Apr/18

  find lim_(x→0^+ )   x e^(−lnx)   .

$$\:\:{find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{x}\:{e}^{−{lnx}} \:\:. \\ $$

Commented by prof Abdo imad last updated on 09/Apr/18

the ch. lnx =−t give  lim_(x→0^+ )    x e^(−lnx)   =lim_(t→+∞ )   e^(−t)  .e^t   = e^0  =1

$${the}\:{ch}.\:{lnx}\:=−{t}\:{give} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{x}\:{e}^{−{lnx}} \:\:={lim}_{{t}\rightarrow+\infty\:} \:\:{e}^{−{t}} \:.{e}^{{t}} \:\:=\:{e}^{\mathrm{0}} \:=\mathrm{1} \\ $$

Answered by kyle_TW last updated on 09/Apr/18

lim_(x→0^+ ) x∙(1/e^(lnx) ) =lim_(x→0^+ ) x∙(1/x) = 1

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}x}\centerdot\frac{\mathrm{1}}{{e}^{{lnx}} }\:=\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}x}\centerdot\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1} \\ $$

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