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

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

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

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

0

Solution steps

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

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^{2}))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

= 2 (- 2 \cdot 0)

Simplify

= 0

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\frac{d}{d x} (\frac{x}{f ( x )})

2 y y^{^{'}} - x y^{^{'}} = y - 2 x

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