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

The limit as x approaches 0+of (ln(x+1))^x is 1

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 (ln(x+1))^x

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

1

Solution steps

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

Apply exponent rules

= x \to 0 + lim (e^{x l n (l n (x + 1))})

Apply the Limit Chain Rule:

1

= 1

x \to 2 lim (\frac{x ^{2} + 5}{x + 1})

x \to 7 - lim (\frac{( x ^{2} - 5 x - 14 )}{x ^{2} - 49})

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

x \to 1 + lim (\frac{1}{x - 1})

x \to 0 lim (\frac{arctan ( x ) - arcsin ( x )}{x ( 1 - cos ( x ) )})

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