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

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

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

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

0

Solution steps

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

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

x^{2} ln^{2} (x) = (x ln (x))^{2}

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

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

= (- 0)^{2}

Simplify

= 0

x \to 0 + lim (x^{2} ln (2 x^{3}))

x \to 0 lim (ln (1 - e^{- x}))

x \to 5 lim (x^{2} - 4 x + 3)

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

x \to 0 lim (1 + \frac{tan ( x )}{x})

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