What is the limit as x approaches 0+of x^2ln(x) ?

The limit as x approaches 0+of x^2ln(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^2ln(x)

x \to 0 + lim (x^{2} ln (x))

0

Solution steps

x \to 0 + lim (x^{2} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x ^{2}}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}})

= x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}})

Simplify

\frac{\frac{1}{x}}{- \frac{2}{x ^{3}}} : - \frac{x ^{2}}{2}

= x \to 0 + lim (- \frac{x ^{2}}{2})

Plug in the value

x = 0

= - \frac{0 ^{2}}{2}

Simplify

- \frac{0 ^{2}}{2} : 0

= 0

x \to (1) lim (\frac{x ^{8} - 1}{x ^{5} - 1})

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

\int 1 - z^{2}

\int \frac{x ^{3} + x}{x - 1} d x

h \to 0 lim (\frac{1}{h})

What is the limit as x approaches 0+of x^2ln(x) ?
The limit as x approaches 0+of x^2ln(x) is 0