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

The limit as x approaches 0 of ln(1+e^{-x}) is ln(2)

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limit as x approaches 0 of ln(1+e^{-x})

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Plug in the value

x = 0

= ln (1 + e^{- 0})

Simplify

= ln (2)

x \to - \infty lim (\frac{a x ^{2} + b x + c}{d - a x})

x \to 0 lim (4 x^{2} + 3 x - 2)

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

x \to 0 + lim (\frac{x}{e ^{- \frac{8}{x}}})

x \to - 1 lim (\frac{x ^{2} + 6}{x + 1})

What is the limit as x approaches 0 of ln(1+e^{-x}) ?
The limit as x approaches 0 of ln(1+e^{-x}) is ln(2)