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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

= x \to 0 + lim (- 2 x)

Plug in the value

x = 0

= - 2 \cdot 0

Simplify

= 0

y = 9 - x^{2}, y = 0, x

\frac{d}{d x} (\frac{( 3 x - 1 )}{x ^{\frac{2}{3}}})

\frac{\partial}{\partial x} (4 + x ln (x y - 5))

\frac{d}{d x} (x (2 - x))

\int \frac{5 x}{e ^{2 x}} d x

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