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

$$find{lim}_{x\to 0^{+}}x{e}^{-lnx}.$$

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

$$thech.lnx=-tgive$$$${lim}_{x\to 0^{+}}x{e}^{-lnx}={lim}_{t\to +\mathrm{\infty }}{e}^{-t}.e^t=e^0=1$$

Answered by kyle_TW last updated on 09/Apr/18

$$\underset{x\to 0^{+}}{limx}\cdot \frac{1}{e^{lnx}}=\underset{x\to 0^{+}}{limx}\cdot \frac{1}{x}=1$$

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