Question

Boa noite pessoal me ajudem nessa de limite para prova amanha
lim de x^3+7x^2+4x+8 / x^3-3x^2+2x quando x tende a -4 ?

Answer 1

$lim_{x \rightarrow -4} \left (\frac{ x^3+7x^2+4x+8}{x^3-3x^2+2x}\right ) =  \left ( \frac{ (-4)^3+7(-4)^2+4(-4)+8}{(-4)^3-3(-4)^2+2(-4)}\right )= \frac{40}{-120}=\frac{-1}{3}$

Answer 2

 Fatoremos o denominador:

$x^3-3x^2+2x=\\x(x^2-3x+2)=\\x(x^2-x-2x+2)=\\x[x(x-1)-2(x-1)]=\\x[(x-1)(x-2)]=\\\boxed{x(x-1)(x-2)}$

 Note que "- 4" não zera o denominador, com isso, podemos substituir tal valor no limite.

 Segue,

$\lim_{x\to-4}\frac{x^3+7x^2+4x+8}{x^3-3x^2+2x}=\\\\\lim_{x\to-4}\frac{x^3\left(1+\frac{7}{x}+\frac{4}{x^2}+\frac{8}{x^3}\right)}{x^3\left(1-\frac{3}{x}+\frac{2}{x^2}\right)}=\\\\\lim_{x\to-4}\frac{1+\frac{7}{x}+\frac{4}{x^2}+\frac{8}{x^3}}{1-\frac{3}{x}+\frac{2}{x^2}}=$

$\frac{1+\frac{7}{-4}+\frac{4}{16}+\frac{8}{-64}}{1-\frac{3}{-4}+\frac{2}{16}}=\\\\\frac{1-\frac{7}{4}+\frac{1}{4}-\frac{1}{8}}{1+\frac{3}{4}+\frac{1}{8}}=\\\\\frac{8-14+2-1}{8}\div\frac{8+6+1}{8}=\\\\\frac{-5}{8}\times\frac{8}{15}=\\\\\boxed{-\frac{1}{3}}$