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

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

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limit as x approaches 0+of 4sqrt(x)ln(x)

x \to 0 + lim (4 x ln (x))

0

Solution steps

x \to 0 + lim (4 x ln (x))

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{2 x ^{\frac{3}{2}}}})

= 4 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{2 x ^{\frac{3}{2}}}})

Simplify

\frac{\frac{1}{x}}{- \frac{1}{2 x ^{\frac{3}{2}}}} : - 2 x^{\frac{1}{2}}

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

Plug in the value

x = 0

= 4 (- 2 \cdot 0^{\frac{1}{2}})

Simplify

= 0

x \to - 3 + lim (\frac{x + 2}{x ^{2} - 9})

x \to - 1 lim (\frac{x}{x})

x \to 0 lim ([\frac{1}{x ^{2}} - \frac{1}{ln ( 1 + x ^{2} )}])

x \to 4 - lim (\frac{a x + b}{x + c})

x \to \infty lim (x^{2} + 3 x + 19 - x)

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