Frederick Kramer

2022-07-04

The question is: $\underset{x\to {0}^{+}}{lim}{x}^{1/x}.$

Hayley Mccarthy

Expert

Let $t=\frac{1}{x}$. When $x\to {0}^{+}$ is going to $+\mathrm{\infty }$. So, the limit is going to become:
$\underset{t\to +\mathrm{\infty }}{lim}{\left(\frac{1}{t}\right)}^{t}=\underset{t\to +\mathrm{\infty }}{lim}{e}^{-\mathrm{ln}\left(t\right)\cdot t}=0$
${a}^{b}={e}^{\mathrm{ln}\left(a\right)\cdot b}$

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