What is the limit as x approaches infinity of ln(x^{4/x}) ?

The limit as x approaches infinity of ln(x^{4/x}) is 0

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limit as x approaches infinity of ln(x^{4/x})

x \to \infty lim (ln (x^{\frac{4}{x}}))

0

Solution steps

x \to \infty lim (ln (x^{\frac{4}{x}}))

ln (x^{\frac{4}{x}}) = \frac{4}{x} ln (x)

= x \to \infty lim (\frac{4}{x} ln (x))

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

= 4 \cdot x \to \infty lim (\frac{1}{x} ln (x))

Simplify

\frac{1}{x} ln (x) : \frac{ln ( x )}{x}

= 4 \cdot x \to \infty lim (\frac{ln ( x )}{x})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{1}{x}}{1})

= 4 \cdot x \to \infty lim (\frac{\frac{1}{x}}{1})

Simplify

= 4 \cdot x \to \infty lim (\frac{1}{x})

Apply Infinity Property:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 4 \cdot 0

Simplify

= 0

\int \frac{x ^{3} + 2}{x ^{2} - 1} d x

y^{^{'}} - \frac{8}{x} y = \frac{y ^{5}}{x ^{11}}

\int_{1}^{2} (2 x + x^{2}) d x

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

x y^{2} \frac{d y}{d x} = y^{3} + x^{3}, y (1) = 5

What is the limit as x approaches infinity of ln(x^{4/x}) ?
The limit as x approaches infinity of ln(x^{4/x}) is 0