What is the limit as x approaches infinity of (ln(x))^{1/(x^2)} ?

The limit as x approaches infinity of (ln(x))^{1/(x^2)} is 1

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limit as x approaches infinity of (ln(x))^{1/(x^2)}

x \to \infty lim ((ln (x))^{\frac{1}{x ^{2}}})

1

Solution steps

x \to \infty lim ((ln (x))^{\frac{1}{x ^{2}}})

Apply exponent rules

= x \to \infty lim (e^{\frac{1}{x ^{2}} l n (l n (x))})

Apply the Limit Chain Rule:

1

= 1

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y^{^{'}} + a^{2} y = 0

\frac{\partial}{\partial x} (- 2 sin (2 x) - 2 x e^{5 y})

f (s) = \frac{s}{s ^{2} + 1}

\int_{- 1}^{0} \frac{3}{3 x - 2} d x

What is the limit as x approaches infinity of (ln(x))^{1/(x^2)} ?
The limit as x approaches infinity of (ln(x))^{1/(x^2)} is 1