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

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

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limit as x approaches 0+of (x^x-1)/x

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

- \infty

Solution steps

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

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

x \to 0 + lim (ln (x) + 1) = - \infty

= 1 \cdot (- \infty)

Apply Infinity Property:

c \cdot (- \infty) = - \infty

= - \infty

\frac{d y}{d x} = sec (x) + y tan (x)

\int \frac{2}{9} x^{- \frac{1}{9}} d x

f (x) = x^{4} - 20 x^{2} + 64

\int (sin^{4} (3 x) cos^{3} (3 x)) d x

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

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