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

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

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

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

1

Solution steps

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

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

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

Simplify

= 1

e^{- 3 x}

(f (x) (x^{2}))^{'}

f (x) = \frac{sin ( x )}{cos ( ln ( - x ) )}

x y - 4 y^{2} = 8

\int 5 x^{3} d x

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