What is the limit as x approaches 0 of 1/(x^2+x)-1/x ?

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

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

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

- 1

Solution steps

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}}))

Simplify

\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})

Plug in the value

x = 0

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

Simplify

= - 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}})

What is the limit as x approaches 0 of 1/(x^2+x)-1/x ?
The limit as x approaches 0 of 1/(x^2+x)-1/x is -1