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

The limit as x approaches 0 of+(xe^{1/x}) is diverges

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+(xe^{1/x})

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

diverges

Solution steps

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

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

then the limit does not exist

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

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

= diverges

x \to \infty lim ((x)^{0})

x \to 0 - lim (csch (x))

x \to 0 - lim (ln (1 + x))

x \to 5 lim (6 x - 14)

x \to 0 + lim ((5 tan (x))^{x})

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