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} \cdot 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

y^{^{''}} + 43 y^{^{'}} + 12 y = 0

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x \to 0 lim (1 - \frac{1}{x})

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