What is the limit as x approaches 1+of (ln(x))^{x-1} ?

The limit as x approaches 1+of (ln(x))^{x-1} is 1

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limit as x approaches 1+of (ln(x))^{x-1}

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

1

Solution steps

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

Apply exponent rules

= x \to 1 + lim (e^{(x - 1) l n (l n (x))})

Apply the Limit Chain Rule:

1

= 1

\frac{d}{d x} (8 x^{3} - 108 x^{2} + 480 x)

x \to 0 lim ((0.5)^{x})

y^{'} = y^{2}

\int \frac{x ( x - 1 )}{1 - x 2 - x} d x

\int yz d x

What is the limit as x approaches 1+of (ln(x))^{x-1} ?
The limit as x approaches 1+of (ln(x))^{x-1} is 1