What is the limit as x approaches 0+of ln(x^x) ?

The limit as x approaches 0+of ln(x^x) is 0

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limit as x approaches 0+of ln(x^x)

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

0

Solution steps

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

ln (x^{x}) = x ln (x)

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}} : - x

= x \to 0 + lim (- x)

Plug in the value

x = 0

= - 0

Simplify

= 0

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What is the limit as x approaches 0+of ln(x^x) ?
The limit as x approaches 0+of ln(x^x) is 0