What is the limit as x approaches 0+of-x^2ln(6x^3) ?

The limit as x approaches 0+of-x^2ln(6x^3) is 0

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limit as x approaches 0+of-x^2ln(6x^3)

x \to 0 + lim (- x^{2} ln (6 x^{3}))

0

Solution steps

x \to 0 + lim (- x^{2} ln (6 x^{3}))

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

= - x \to 0 + lim (x^{2} ln (6 x^{3}))

Rewrite for L'Hopital

= - x \to 0 + lim (\frac{ln ( 6 x ^{3} )}{\frac{1}{x ^{2}}})

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

= - (- \frac{3 \cdot 0 ^{2}}{2})

Simplify

= 0

\int \frac{6}{5 + 6 x} d x

\int e^{x} 2 d x

\frac{\partial}{\partial x} (6 e^{x y + 5})

ln (\frac{( 1 + x )}{1 - x})

\frac{d}{d x} (- (x - 3)^{2} (1 - x)^{2})

What is the limit as x approaches 0+of-x^2ln(6x^3) ?
The limit as x approaches 0+of-x^2ln(6x^3) is 0