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

Popular >

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

x \to 0 lim (\frac{2 x}{x - 2})

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

x \to 3 lim (\frac{3 + x}{x ^{2} - 9})

x \to - 1 lim (4 x^{2} - 5 x)

x \to 0 - lim (x^{\frac{1}{2}})

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