Question

Se puder resolver com cálculo seria ótimo...
Desde já obrigada...

Answer 1

Existe uma fatoração, não muito vista, que nos ajudará a resolver esse limite:

$\boxed{\boxed{a^{3}+b^{3}=(a+b)\cdot(a^{2}-ab+b^{2})}}$
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$\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=\lim\limits_{x\rightarrow-2}~\dfrac{2^{3}+x^{3}}{2^{2}-x^{2}}$

Fatorando o numerador e o denominador:

$\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=\lim\limits_{x\rightarrow-2}~\dfrac{(2+x)\cdot(2^{2}-2x+x^{2})}{(2+x)\cdot(2-x)}\\\\\\\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=\lim\limits_{x\rightarrow-2}~\dfrac{4-2x+x^{2}}{2-x}\\\\\\\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=\dfrac{4-2(-2)+(-2)^{2}}{2-(-2)}\\\\\\\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=\dfrac{4+4+4}{2+2}=\dfrac{12}{4}\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow-2}~\dfrac{8+x^{3}}{4-x^{2}}=3}}$