Question

lim 8+x³/4-x² quando x tende a -2

Answer 1

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Solução!

Antes de resolver esses limites tem que fatorar os termos.

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

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Answer 2

$\lim_{n \to -2} \frac{8+x^3}{4-x^2} = \\ \\ = \lim_{n \to -2} \frac{(x+2)(x^2-2x+4)}{-(x-2)(x+2)} = \\ \\ =\lim_{n \to -2} -(\frac{x^2-2x+4}{x-2} )= \\ \\ =- \frac{ \lim_{n \to-2} (x^2-2x+4) }{ \lim_{n \to -2} (x-2) } = \\ \\=- \frac{(-2)^2-2.(-2)+4}{-2-2} = \\ \\= -\frac{4+4+4}{-4} =- \frac{12}{-4} =3$