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

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

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

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

\frac{1}{e}

Decimal

0.36787 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= \frac{1}{e}

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