What is the limit as x approaches 0+of x^{14}ln(x) ?

The limit as x approaches 0+of x^{14}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^{14}ln(x)

x \to 0 + lim (x^{14} ln (x))

0

Solution steps

x \to 0 + lim (x^{14} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{14}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{14}{x ^{15}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{14}{x ^{15}}})

Simplify

\frac{\frac{1}{x}}{- \frac{14}{x ^{15}}} : - \frac{x ^{14}}{14}

= x \to 0 + lim (- \frac{x ^{14}}{14})

Plug in the value

x = 0

= - \frac{0 ^{14}}{14}

Simplify

- \frac{0 ^{14}}{14} : 0

= 0

f (x) = - sin (2 x) \cdot 2

\frac{\partial}{\partial y} (cos (2 x - y))

x \to 1 lim (x^{2} - 4 x + 3)

\int \frac{x}{e ^{- x^{2}}} d x

\frac{\partial}{\partial x} (2 x ln (y))

What is the limit as x approaches 0+of x^{14}ln(x) ?
The limit as x approaches 0+of x^{14}ln(x) is 0