What is the limit as x approaches 0+of \sqrt[5]{x}ln(x) ?

The limit as x approaches 0+of \sqrt[5]{x}ln(x) is 0

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limit as x approaches 0+of \sqrt[5]{x}ln(x)

0

Solution steps

Rewrite for L'Hopital

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{5 x ^{\frac{6}{5}}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{5 x ^{\frac{6}{5}}}})

Simplify

\frac{\frac{1}{x}}{- \frac{1}{5 x ^{\frac{6}{5}}}} : - 5 x^{\frac{1}{5}}

= x \to 0 + lim (- 5 x^{\frac{1}{5}})

Plug in the value

x = 0

= - 5 \cdot 0^{\frac{1}{5}}

Simplify

= 0

x \to \infty lim (x e^{- x} ln (x))

x \to 0 lim (3 x^{4} - 2 x - 1)

x \to 0 + lim (x tan (\frac{π}{2} - x))

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

x \to 4 + lim (x^{2} + 5 x - 35)

What is the limit as x approaches 0+of \sqrt[5]{x}ln(x) ?
The limit as x approaches 0+of \sqrt[5]{x}ln(x) is 0