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

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

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}))

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

= - 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}}})

Simplify

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

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

Plug in the value

x = 0

= - (- 0)

Simplify

= 0

x \to 3 lim ((3 x - 10) + x)

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

x \to \frac{1}{3} lim ((2 - 3 x)^{2} + 1)

x \to - 5 lim (x^{2} - 6)

x \to \frac{5}{3} lim (x^{3} - 2 x + 3)

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