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

解答

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

- 1

求解步骤

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

\frac{1}{x ^{2} + x} - \frac{1}{x} = \frac{1}{x ^{2} + x} (1 - \frac{\frac{1}{x}}{\frac{1}{x ^{2} + x}})

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

化简

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

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

代入值

x = 0

= - \frac{1}{0 + 1}

化简

= - 1

x \to - 5 lim (\frac{( x ^{2} + 6 x + 5 )}{x ^{2} + 4 x - 5})

y^{^{'}} = 3 y + 2

\int y (x + y) d y

\int \frac{1}{( 121 - x ^{2} ) ^{\frac{3}{2}}} d x

x \to 0 + lim (x^{\frac{8}{x}})