limit as h approaches 0 of (ln(2e+h)-ln(2e))/h

解答

h \to 0 lim (\frac{ln ( 2 e + h ) - ln ( 2 e )}{h})

\frac{1}{2 e}

十进制

0.18393 \dots

求解步骤

h \to 0 lim (\frac{ln ( 2 e + h ) - ln ( 2 e )}{h})

使用洛必达法则

= h \to 0 lim (\frac{\frac{1}{h + 2 e}}{1})

代入值

h = 0

= \frac{\frac{1}{0 + 2 e}}{1}

化简

= \frac{1}{2 e}

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

x \to 1 lim (2 x - 4)

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

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

x \to 4 - lim (x - 2)