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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

= 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}}}})

= 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

y = \frac{1 - y}{x - 1}

x \to + 7 lim (2 cos (x))

\frac{d}{d x} (2 x^{2} + 4 x + 7)

\frac{\partial}{\partial x} (5 x^{2} + 3 y^{2})

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