What is the limit as z approaches 2 of ((2z-z^2)/(z^2-4))^3 ?

The limit as z approaches 2 of ((2z-z^2)/(z^2-4))^3 is -1/8

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limit as z approaches 2 of ((2z-z^2)/(z^2-4))^3

z \to 2 lim ((\frac{2 z - z ^{2}}{z ^{2} - 4})^{3})

- \frac{1}{8}

Decimal Notation

- 0.125

Solution steps

z \to 2 lim ((\frac{2 z - z ^{2}}{z ^{2} - 4})^{3})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= (z \to 2 lim (\frac{2 z - z ^{2}}{z ^{2} - 4}))^{3}

Simplify

\frac{2 z - z ^{2}}{z ^{2} - 4} : - \frac{z}{z + 2}

= (z \to 2 lim (- \frac{z}{z + 2}))^{3}

Plug in the value

z = 2

= (- \frac{2}{2 + 2})^{3}

Simplify

(- \frac{2}{2 + 2})^{3} : - \frac{1}{8}

= - \frac{1}{8}

x \to - 2 lim (4 x^{2} + 3 x - 2)

x \to 3 - lim (\frac{4}{x - 3})

x \to 0 lim (cos (2 x) - 1)

x \to 1.5 lim (\frac{7 x ^{2} + 14 x - 21}{3 - 2 x})

x \to - 1 lim (5 x^{2} - 3 x - 2)

What is the limit as z approaches 2 of ((2z-z^2)/(z^2-4))^3 ?
The limit as z approaches 2 of ((2z-z^2)/(z^2-4))^3 is -1/8