What is the limit as x approaches 2 of (x^2+8)(2x-4) ?

The limit as x approaches 2 of (x^2+8)(2x-4) is 0

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limit as x approaches 2 of (x^2+8)(2x-4)

x \to 2 lim ((x^{2} + 8) (2 x - 4))

0

Solution steps

x \to 2 lim ((x^{2} + 8) (2 x - 4))

Plug in the value

x = 2

= (2^{2} + 8) (2 \cdot 2 - 4)

Simplify

= 0

y \to - \infty lim (\frac{2 - y}{7 + 6 y ^{2}})

\frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

\int_{0}^{2} (4 - x^{2})^{2} d x

(1 + x^{2}) \frac{d y}{d x} + 4 x y = (1 + x^{2})^{- 2}

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

What is the limit as x approaches 2 of (x^2+8)(2x-4) ?
The limit as x approaches 2 of (x^2+8)(2x-4) is 0