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 Notation

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}

x \to 2 lim (\frac{1}{x} - \frac{1}{x ^{3}})

z \to 0 lim (\frac{z}{sin ( z )})

x \to - 2 lim ((3 x + 8)^{x + 1})

x \to 0 lim (\frac{x}{∣ x ∣})

a \to 0 lim (a \cdot sin (x + \frac{π}{a}))

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