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

解答

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

e

十进制

2.71828 \dots

求解步骤

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

使用洛必达法则

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

代入值

x = 0

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

化简

= e

x \to - \infty lim (16 x^{3})

t \to 0 lim (\frac{2}{t} - \frac{1}{e ^{t} - 1})

x \to 0 lim (\frac{- 2 x ^{2}}{x ^{2}})

x \to 0 + lim \frac{( ( ln ( x ) ) ^{2} )}{ln ( sin ( x ) )}

x \to \infty lim (3^{x} x!)