What is the limit as x approaches 3 of (x^3-2x^2-4x+8)/(3x^2+3x-6) ?

The limit as x approaches 3 of (x^3-2x^2-4x+8)/(3x^2+3x-6) is 1/6

limit as x approaches 3 of (x^3-2x^2-4x+8)/(3x^2+3x-6)

x \to 3 lim (\frac{x ^{3} - 2 x ^{2} - 4 x + 8}{3 x ^{2} + 3 x - 6})

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

x \to 3 lim (\frac{x ^{3} - 2 x ^{2} - 4 x + 8}{3 x ^{2} + 3 x - 6})

Plug in the value

x = 3

= \frac{3 ^{3} - 2 \cdot 3 ^{2} - 4 \cdot 3 + 8}{3 \cdot 3 ^{2} + 3 \cdot 3 - 6}

Simplify

\frac{3 ^{3} - 2 \cdot 3 ^{2} - 4 \cdot 3 + 8}{3 \cdot 3 ^{2} + 3 \cdot 3 - 6} : \frac{1}{6}

= \frac{1}{6}

x \to \infty lim (\frac{3 x ^{3} - 4 x ^{2} + 2}{x ^{3} + 2 x + 7})

x \to 3 lim (3 x + 4)

x \to 0 lim (\frac{x}{1 + 3 ^{\frac{1}{x}}})

x \to \infty lim ((\frac{3 n + 1}{5 + 3 n})^{n^{2} + 2})

x \to 3 lim (3 x - 8)

What is the limit as x approaches 3 of (x^3-2x^2-4x+8)/(3x^2+3x-6) ?
The limit as x approaches 3 of (x^3-2x^2-4x+8)/(3x^2+3x-6) is 1/6