What is the limit as x approaches 0+of x^{18}ln(x) ?

The limit as x approaches 0+of x^{18}ln(x) is 0

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limit as x approaches 0+of x^{18}ln(x)

x \to 0 + lim (x^{18} ln (x))

0

Solution steps

x \to 0 + lim (x^{18} ln (x))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{18}{x ^{19}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{18}{x ^{19}}})

Simplify

\frac{\frac{1}{x}}{- \frac{18}{x ^{19}}} : - \frac{x ^{18}}{18}

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

Plug in the value

x = 0

= - \frac{0 ^{18}}{18}

Simplify

- \frac{0 ^{18}}{18} : 0

= 0

x \to 3 - lim (\frac{1}{x - 3})

f (x) = \frac{5}{3 x}

\int (- 0.07 x) d x

\int x - e^{- x} d x

\frac{d}{d x} (\frac{9}{3 x})

What is the limit as x approaches 0+of x^{18}ln(x) ?
The limit as x approaches 0+of x^{18}ln(x) is 0