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

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

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^2-x-ln(x)

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

\infty

Solution steps

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

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

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

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

x \to 0 + lim (x) = 0

x \to 0 + lim (ln (x)) = - \infty

= 0 - 0 - (- \infty)

Simplify

0 - 0 - (- \infty) : \infty

= \infty

\int \frac{x ^{3} - 2 x - 2}{x ^{4} + 2 x ^{3} + 3 x ^{2}} d x

\frac{d}{d x} (x^{\frac{4}{5}} (x - 6)^{2})

\frac{d}{d x} (1 + cot^{2} (x))

x \to 0 lim (\frac{x}{- 2 x ^{4} - 14 x})

x \to - 2 lim (\frac{x ^{2}}{x ^{2} - 4})

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