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

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

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

1

Solution steps

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

\frac{ln ( 1 + x )}{x} = ln (1 + x) \frac{1}{x}

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

Simplify

ln (1 + x) \frac{1}{x} : \frac{ln ( 1 + x )}{x}

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{\frac{1}{1 + 0}}{1}

Simplify

= 1

x \to 0 + lim (1^{\frac{l n ( 6 )}{x}})

x \to 0 lim ((1 + 8 x)^{\frac{1}{x}})

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

x \to - 3 lim (\frac{4 - x ^{2}}{x + 3})

θ \to π lim (θ sec (θ))

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