What is the limit as x approaches 0+of 4x^2ln(x) ?

The limit as x approaches 0+of 4x^2ln(x) is 0

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limit as x approaches 0+of 4x^2ln(x)

x \to 0 + lim (4 x^{2} ln (x))

0

Solution steps

x \to 0 + lim (4 x^{2} ln (x))

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

= 4 \cdot x \to 0 + lim (x^{2} ln (x))

Rewrite for L'Hopital

= 4 \cdot 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}}})

= 4 \cdot 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}

= 4 \cdot x \to 0 + lim (- \frac{x ^{2}}{2})

Plug in the value

x = 0

= 4 (- \frac{0 ^{2}}{2})

Simplify

4 (- \frac{0 ^{2}}{2}) : 0

= 0

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

x \to - \infty lim (1 0^{x} + 2)

x \to \infty lim (5 + \frac{8}{x})

x \to 7 - lim (\frac{5 - x}{x - 7})

x \to 3 lim (x^{2} + 5 x - 7)

What is the limit as x approaches 0+of 4x^2ln(x) ?
The limit as x approaches 0+of 4x^2ln(x) is 0