What is the limit as x approaches 0+of x^{1/2}*ln(x) ?

The limit as x approaches 0+of x^{1/2}*ln(x) is 0

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limit as x approaches 0+of x^{1/2}*ln(x)

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

0

Solution steps

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

Rewrite for L'Hopital

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

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^{^{'}} = 4 x + y

x \to 5 lim (x (x - 2))

\int \frac{( x + 8 ) ^{6}}{2 x} d x

6 - 6 x

\int (x^{3}) d x

What is the limit as x approaches 0+of x^{1/2}*ln(x) ?
The limit as x approaches 0+of x^{1/2}*ln(x) is 0