What is the limit as x approaches-1-of 2x^2-4x+8 ?

The limit as x approaches-1-of 2x^2-4x+8 is 14

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limit as x approaches-1-of 2x^2-4x+8

x \to - 1 - lim (2 x^{2} - 4 x + 8)

14

Solution steps

x \to - 1 - lim (2 x^{2} - 4 x + 8)

Plug in the value

x = - 1

= 2 (- 1)^{2} - 4 (- 1) + 8

Simplify

2 (- 1)^{2} - 4 (- 1) + 8 : 14

= 14

\frac{\partial}{\partial x} (6 x y + 5 yz + x z - 48)

16 y^{^{''}} + 152 y^{^{'}} + 361 y = 0, y (0) = 9, y^{^{'}} (0) = 2

\frac{\partial}{\partial y} (ln (1 + x^{2} + y))

\int \frac{18}{( x ^{2} - 1 ) ^{2}} d x

y^{^{'}} + 7 x^{- 1} y = x^{2}

What is the limit as x approaches-1-of 2x^2-4x+8 ?
The limit as x approaches-1-of 2x^2-4x+8 is 14