limit as x approaches-infinity of (3e^x-4e^{-x})/(2e^x+e^{-x)}

解

x \to - \infty lim (\frac{3 e ^{x} - 4 e ^{- x}}{2 e ^{x} + e ^{- x}})

- 4

解答ステップ

x \to - \infty lim (\frac{3 e ^{x} - 4 e ^{- x}}{2 e ^{x} + e ^{- x}})

簡素化

\frac{3 e ^{x} - 4 e ^{- x}}{2 e ^{x} + e ^{- x}} : \frac{3 e ^{2 x} - 4}{2 e ^{2 x} + 1}

= x \to - \infty lim (\frac{3 e ^{2 x} - 4}{2 e ^{2 x} + 1})

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

不定形式の場合を除く

= \frac{lim _{x \to - \infty} ( 3 e ^{2 x} - 4 )}{lim _{x \to - \infty} ( 2 e ^{2 x} + 1 )}

x \to - \infty lim (3 e^{2 x} - 4) = - 4

x \to - \infty lim (2 e^{2 x} + 1) = 1

= \frac{- 4}{1}

簡素化

= - 4