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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{d}{d x} (ln (x^{2} + 3 x + 3))

f (x) = 4 x^{4} + 3 x^{2} - 1

\frac{d y}{d x} = 4 y x

\int cos^{7} (x) d x

\frac{\partial}{\partial x} (ln (4 + x^{2} 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