What is the limit as x approaches 0+of x^4ln(x) ?

The limit as x approaches 0+of x^4ln(x) is 0

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limit as x approaches 0+of x^4ln(x)

x \to 0 + lim (x^{4} ln (x))

0

Solution steps

x \to 0 + lim (x^{4} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{4}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{4}{x ^{5}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{4}{x ^{5}}})

Simplify

\frac{\frac{1}{x}}{- \frac{4}{x ^{5}}} : - \frac{x ^{4}}{4}

= x \to 0 + lim (- \frac{x ^{4}}{4})

Plug in the value

x = 0

= - \frac{0 ^{4}}{4}

Simplify

- \frac{0 ^{4}}{4} : 0

= 0

2 cos (2 y)

\int_{0}^{2} 4 x^{2} d x

\frac{\partial}{\partial x} (ln (1 - 2 x))

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

\frac{e ^{2 x + 1}}{2 x + 5}

What is the limit as x approaches 0+of x^4ln(x) ?
The limit as x approaches 0+of x^4ln(x) is 0