Evaluate: lim\(_{x→-2}\) \(\frac{x^3+8}{x+2}\).

Explanation

\(\frac{x^3+8}{x+2}\).

x\(^3\) + 8 = x\(^3\) + 23
recall that
x\(^3\) + y\(^3\) = (x + y)(x\(^2\) - xy + y\(^2\))
x = x, y = 2
x\(^3\) + 8 = (x + 2)(x\(^3\) - 2(x) + 2\(^2\))

\(\frac{x^3 + 8}{x + 2} = \frac{( x + 2)(x^3 - 2(x) + 2^2))}{x+2}\) → x\(^2\) - 2x + 4

x = -2
(-2)\(^2\) - 2(-2) + 4 = 4 + 4 + 4

= 12