What is the limit as x approaches 0+of x^2(ln(x)) ?

The limit as x approaches 0+of x^2(ln(x)) is 0

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limit as x approaches 0+of x^2(ln(x))

x \to 0 + lim (x^{2} (ln (x)))

0

Solution steps

x \to 0 + lim (x^{2} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{2}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}})

Simplify

\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}} : - \frac{x ^{2}}{2}

= x \to 0 + lim (- \frac{x ^{2}}{2})

Plug in the value

x = 0

= - \frac{0 ^{2}}{2}

Simplify

- \frac{0 ^{2}}{2} : 0

= 0

\frac{d}{d x} (sec (x - \frac{π}{4}))

\frac{\partial}{\partial x} (7 e^{x y + 5})

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

(- \frac{10}{63}, \frac{1}{3}), (\frac{2}{3}, - \frac{11}{45})

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What is the limit as x approaches 0+of x^2(ln(x)) ?
The limit as x approaches 0+of x^2(ln(x)) is 0