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

解答

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

1

求解步骤

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

化简

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

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

改写以应用洛必达法则

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

使用洛必达法则

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

化简

\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

不定型除外

= \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}

化简

= 1

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

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

d er i v a t i v e - \frac{5}{x ^{6}}

d er i v a t i v e ln (2^{x})

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