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

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

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

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

1

Solution steps

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

Apply exponent rules

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

Apply the Limit Chain Rule:

1

= 1

\int (x - 1) 1 + e^{2 y} d y

\int (\frac{1}{2 x + 5}) d x

5 x^{^{''}} + 50 x = 10

\int x (5 x^{2} + \frac{2}{3})^{4} d x

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

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