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 Does not exist

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

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

Does not exist

Solution steps

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

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

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

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

Does not exist

= Does not exist

\int x \cdot (2 x^{2} + 1)^{\frac{3}{2}} d x

\frac{d y}{d x} - \frac{6}{x} y = \frac{y ^{4}}{x ^{15}}, y (1) = 1

x \frac{d y}{d x} = - 4 y + x^{3} - x

\frac{d x}{d t} = 4 x (9 - x)

(3 x - 2)^{3} (x^{3} + 1)^{2}

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 Does not exist