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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

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

Simplify

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

= 0

x \to \infty lim (3^{x} x)

x \to 2 lim (\frac{x + 5}{x - 4})

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

x \to 4 lim (\frac{3 x ^{2} - 12 x}{x - 2})

x \to 4 lim (3 x^{2} - 2 x)

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