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

The limit as h approaches 0 of (ln(1+h)-0)/h is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as h approaches 0 of (ln(1+h)-0)/h

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

1

Solution steps

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

Simplify

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

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

Plug in the value

h = 0

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

Simplify

= 1

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