Question

Lim x^3+8/x+2 quando x tende a menos 2

Answer 1

$\displaystyle\lim_{x \to -2} \dfrac{x^3+8}{x+2} = \displaystyle\lim_{x \to -2} \dfrac{(-2)^3+8}{-2+2} = \dfrac{-8+8}{-2+2}=\dfrac{0}{0} \\ \\ \displaystyle\lim_{x \to -2} \dfrac{x^3+8}{x+2} = \displaystyle\lim_{x \to -2} \dfrac{x^3+2^3}{x+2}= \\ \\ \displaystyle\lim_{x \to -2} \dfrac{(x+2)(x^2-2x+4)}{x+2}=\displaystyle\lim_{x \to -2} x^2-2x+4 = 12$

Answer 2

(x+2)³=x³+3*x²*2+3*x*2²+2³

(x+2)³=x³+2³+3*x²*2+3*x*2²

(x+2)³=x³+2³+3*x*2*(x+2)

x³+2³= (x+2)³- 6x*(x+2)

x³+2³= (x+2)*[(x+2)²-6x]

x³+8= (x+2)*[x²+4x+4-6x]

x³+8= (x+2)*(x²+4-2x)

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Lim (x³+8)/(x+2)

x-->-2

Lim (x+2)*(x²+4-2x)/(x+2)

x-->-2

Lim (x²+4-2x) =(-2)²+4 -2*(-2) =4 +4+4=12

x-->-2