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

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

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

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

1

Solution steps

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

Simplify

\frac{ln ( 1 + e ^{- x} )}{e ^{- x}} : e^{x} ln (1 + e^{- x})

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

Rewrite for L'Hopital

= x \to \infty lim (\frac{ln ( 1 + e ^{- x} )}{\frac{1}{e ^{x}}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{- \frac{e ^{- x}}{1 + e ^{- x}}}{- e ^{- x}})

= x \to \infty lim (\frac{- \frac{e ^{- x}}{1 + e ^{- x}}}{- e ^{- x}})

Simplify

\frac{- \frac{e ^{- x}}{1 + e ^{- x}}}{- e ^{- x}} : \frac{1}{1 + e ^{- x}}

= x \to \infty lim (\frac{1}{1 + e ^{- x}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 1 + e ^{- x} )}

x \to \infty lim (1) = 1

x \to \infty lim (1 + e^{- x}) = 1

= \frac{1}{1}

Simplify

= 1

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

\int \frac{3 sin ( x ) + 2 sec ( x )}{tan ( x )} d x

- \frac{5}{x ^{6}}

ln (2^{x})

\int \frac{t ^{2}}{1 + t ^{2}} d t

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