What is the limit as x approaches 0+of x^2(1+ln(x)) ?

The limit as x approaches 0+of x^2(1+ln(x)) is 0

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limit as x approaches 0+of x^2(1+ln(x))

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

0

Solution steps

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

Rewrite for L'Hopital

= x \to 0 + lim (\frac{1 + 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

f (x) = \frac{2}{x ^{3}}

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y^{^{''}} + 4 y = 2 sec (2 x)

\int e^{8 x + 6} d x

What is the limit as x approaches 0+of x^2(1+ln(x)) ?
The limit as x approaches 0+of x^2(1+ln(x)) is 0