What is the limit as x approaches 0+of ln(x^{1/x}) ?

The limit as x approaches 0+of ln(x^{1/x}) is -infinity

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limit as x approaches 0+of ln(x^{1/x})

x \to 0 + lim (ln (x^{\frac{1}{x}}))

- \infty

Solution steps

x \to 0 + lim (ln (x^{\frac{1}{x}}))

ln (x^{\frac{1}{x}}) = \frac{1}{x} ln (x)

= x \to 0 + lim (\frac{1}{x} ln (x))

x \to a lim [f (x) \cdot g (x)] = x \to a lim f (x) \cdot x \to a lim g (x)

With the exception of indeterminate form

= x \to 0 + lim (\frac{1}{x}) \cdot x \to 0 + lim (ln (x))

x \to 0 + lim (\frac{1}{x}) = \infty

x \to 0 + lim (ln (x)) = - \infty

= \infty \cdot (- \infty)

Apply Infinity Property:

\infty (- \infty) = - \infty

= - \infty

What is the limit as x approaches 0+of ln(x^{1/x}) ?
The limit as x approaches 0+of ln(x^{1/x}) is -infinity