What is the limit as x approaches 0+of (x^x-1)/(2x) ?

The limit as x approaches 0+of (x^x-1)/(2x) is -infinity

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limit as x approaches 0+of (x^x-1)/(2x)

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

- \infty

Solution steps

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

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

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

Apply L'Hopital's Rule

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

Simplify

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 1 \cdot (- \infty)

Apply Infinity Property:

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

= - \infty

What is the limit as x approaches 0+of (x^x-1)/(2x) ?
The limit as x approaches 0+of (x^x-1)/(2x) is -infinity