What is the limit as x approaches+0+of x^{12}ln(x) ?

The limit as x approaches+0+of x^{12}ln(x) is 0

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limit as x approaches+0+of x^{12}ln(x)

x \to + 0 + lim (x^{12} ln (x))

0

Solution steps

x \to 0 + lim (x^{12} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{12}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{12}{x ^{13}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{12}{x ^{13}}})

Simplify

\frac{\frac{1}{x}}{- \frac{12}{x ^{13}}} : - \frac{x ^{12}}{12}

= x \to 0 + lim (- \frac{x ^{12}}{12})

Plug in the value

x = 0

= - \frac{0 ^{12}}{12}

Simplify

- \frac{0 ^{12}}{12} : 0

= 0

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y (x - b) + y^{^{'}} b x = 0

y^{^{'}} = \frac{- y ^{3} + 14 x - 8}{3 x y ^{2}}

What is the limit as x approaches+0+of x^{12}ln(x) ?
The limit as x approaches+0+of x^{12}ln(x) is 0