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

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

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

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

0

Solution steps

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

Apply exponent rule:

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

x^{2} (ln (x))^{2} = (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 (\frac{- 2}{x ^{2}})

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x \to 0 lim (∣ x ∣ cos (\frac{π}{x}))

x \to 0 - lim (\frac{9 sin ( x )}{x})

θ \to \frac{π}{6} lim (\frac{sin ( 2 θ )}{sin ( θ ) tan ( θ )})

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