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

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

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

0

Solution steps

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

ln (x^{2}) = 2 ln (x)

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Rewrite for L'Hopital

= 2 \cdot 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}}})

= 2 \cdot 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}

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

Plug in the value

x = 0

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

Simplify

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

= 0

\int (4 x - x^{2})^{2} d x

e^{t} cos (2 t)

- 1 {\frac{15}{s ^{2} + 4 s + 13}}

y = e^{x + 1} + 1

x \frac{d y}{d x} + 2 y = \frac{1}{x ^{3}}

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