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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

= - \frac{0 ^{2} ( 2 \cdot 0 + 1 )}{2 ( 0 + 1 )}

Simplify

- \frac{0 ^{2} ( 2 \cdot 0 + 1 )}{2 ( 0 + 1 )} : 0

= 0

x \to 10 lim (\frac{x ^{3}}{x - 10})

x \to 2 lim ((4 x - 6)^{4})

x \to 3 - lim (\frac{x ^{2} - 9}{x + 3})

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

x \to - 4 + lim (16 - x^{2})

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