What is the limit as x approaches 0 of 1/(2x^2+x) ?

The limit as x approaches 0 of 1/(2x^2+x) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 0 of 1/(2x^2+x)

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

diverges

Solution steps

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

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

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

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

= diverges

What is the limit as x approaches 0 of 1/(2x^2+x) ?
The limit as x approaches 0 of 1/(2x^2+x) is diverges