What is the limit as x approaches 0+of sqrt(x)ln(6x) ?

The limit as x approaches 0+of sqrt(x)ln(6x) is 0

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limit as x approaches 0+of sqrt(x)ln(6x)

x \to 0 + lim (x ln (6 x))

0

Solution steps

x \to 0 + lim (x ln (6 x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( 6 x )}{\frac{1}{x}})

Apply L'Hopital's Rule

= 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}}

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

Plug in the value

x = 0

= - 2 \cdot 0^{\frac{1}{2}}

Simplify

= 0

What is the limit as x approaches 0+of sqrt(x)ln(6x) ?
The limit as x approaches 0+of sqrt(x)ln(6x) is 0