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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

0

Solution steps

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

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