limit as x approaches 0 of 1/x-1/((x+x^2))+ln(e^2+x)

解答

x \to 0 lim (\frac{1}{x} - \frac{1}{( x + x ^{2} )} + ln (e^{2} + x))

3

求解步骤

x \to 0 lim (\frac{1}{x} - \frac{1}{x + x ^{2}} + ln (e^{2} + x))

乘开

\frac{1}{x} - \frac{1}{x + x ^{2}} + ln (e^{2} + x) : \frac{1 + x ln ( x + e ^{2} ) + ln ( x + e ^{2} )}{x + 1}

= x \to 0 lim (\frac{1 + x ln ( x + e ^{2} ) + ln ( x + e ^{2} )}{x + 1})

代入值

x = 0

= \frac{1 + 0 \cdot ln ( 0 + e ^{2} ) + ln ( 0 + e ^{2} )}{0 + 1}

化简

\frac{1 + 0 \cdot ln ( 0 + e ^{2} ) + ln ( 0 + e ^{2} )}{0 + 1} : 3

= 3

x \to 2 lim (\frac{e ^{x} sin ( x )}{( x - 2 ) ^{3}})

x \to 0 lim (ln (1 - cos (x)))

x \to 0 lim (\frac{1}{x} (\frac{1}{x + 4.25} - \frac{1}{4.25}))

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

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