What is the limit as x approaches 0+of x^2*ln(4x^2) ?

The limit as x approaches 0+of x^2*ln(4x^2) is 0

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limit as x approaches 0+of x^2*ln(4x^2)

x \to 0 + lim (x^{2} \cdot ln (4 x^{2}))

0

Solution steps

x \to 0 + lim (x^{2} ln (4 x^{2}))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( 4 x ^{2} )}{\frac{1}{x ^{2}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{2}{x}}{- \frac{2}{x ^{3}}})

= x \to 0 + lim (\frac{\frac{2}{x}}{- \frac{2}{x ^{3}}})

Simplify

\frac{\frac{2}{x}}{- \frac{2}{x ^{3}}} : - x^{2}

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

Plug in the value

x = 0

= - 0^{2}

Simplify

= 0

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

t \to 0 lim (\frac{2 sin ( 4 t )}{t})

x \to \infty lim (3 x^{2} - 1)

x \to 4 lim (5 x - 7)

w \to 0 lim (csc (w) - cot (w))

What is the limit as x approaches 0+of x^2*ln(4x^2) ?
The limit as x approaches 0+of x^2*ln(4x^2) is 0