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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{2}{x}}{- \frac{1}{x ^{2}}} : - 2 x

= x \to 0 + lim (- 2 x)

Plug in the value

x = 0

= - 2 \cdot 0

Simplify

= 0

\frac{d}{d x} (\frac{3 sin ( x ) - 2 cos ( x )}{5})

\int (cos (x))^{2} sin (x) d x

4 x^{2} + x y + 4 y^{2} = 9, (1, 1)

\frac{d}{d x} (3 x cot (4 x - 2))

y = x^{2} - 1, y = 8

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