What is the limit as x approaches 0+of x^{17}ln(x) ?

The limit as x approaches 0+of x^{17}ln(x) is 0

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limit as x approaches 0+of x^{17}ln(x)

x \to 0 + lim (x^{17} ln (x))

0

Solution steps

x \to 0 + lim (x^{17} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{17}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{17}{x ^{18}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{17}{x ^{18}}})

Simplify

\frac{\frac{1}{x}}{- \frac{17}{x ^{18}}} : - \frac{x ^{17}}{17}

= x \to 0 + lim (- \frac{x ^{17}}{17})

Plug in the value

x = 0

= - \frac{0 ^{17}}{17}

Simplify

- \frac{0 ^{17}}{17} : 0

= 0

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What is the limit as x approaches 0+of x^{17}ln(x) ?
The limit as x approaches 0+of x^{17}ln(x) is 0