What is the limit as y approaches 0 of 8/(y^2+y)-8/y ?

The limit as y approaches 0 of 8/(y^2+y)-8/y is -8

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limit as y approaches 0 of 8/(y^2+y)-8/y

y \to 0 lim (\frac{8}{y ^{2} + y} - \frac{8}{y})

- 8

Solution steps

y \to 0 lim (\frac{8}{y ^{2} + y} - \frac{8}{y})

Simplify

\frac{8}{y ^{2} + y} - \frac{8}{y} : - \frac{8}{y + 1}

= y \to 0 lim (- \frac{8}{y + 1})

Plug in the value

y = 0

= - \frac{8}{0 + 1}

Simplify

= - 8

y^{^{''}} + 3 y^{^{'}} - 10 y = x (e^{x} + 1)

\frac{d}{d x} (3^{- 6 x + 6})

f^{^{'}} (x) = 3 x^{2} (8 - x^{- 3})

\int_{0}^{1} 5^{2 x} d x

\int_{0}^{1} x^{3} + x d x

What is the limit as y approaches 0 of 8/(y^2+y)-8/y ?
The limit as y approaches 0 of 8/(y^2+y)-8/y is -8