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

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

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

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

8

Solution steps

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

Plug in the value

x = 2

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

Simplify

2 \cdot 2^{2} - 4 \cdot 2 + 8 : 8

= 8

x \to 4 - lim (- \frac{1}{x ^{2} - 16})

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

x \to 0 lim (\frac{( x - 2 )}{x ^{4}})

x \to 2 lim (\frac{[ x ]}{x})

x \to 1 lim (3 x^{2} + 4 x)

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