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

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

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

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

\frac{1}{e}

Decimal

0.36787 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

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

Simplify

= \frac{1}{e}

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