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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

Simplify

\frac{\frac{2 l n ( 4 x )}{x}}{- \frac{1}{x ^{2}}} : - 2 x ln (4 x)

= x \to 0 + lim (- 2 x ln (4 x))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - 2 \cdot x \to 0 + lim (x ln (4 x))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

Simplify

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

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

Plug in the value

x = 0

= - 2 (- 0)

Simplify

= 0

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{x}

\frac{d}{d x} (9 x^{2} - 2 x + 1)

\frac{d}{d x} (x a^{2} + x^{2})

a re a x = y^{2} - 1, x = 1 - y^{2}

in v erse l a pl a ce \frac{2}{s ^{2} ( s + 2 )}

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