Question

Determine o valor do limite x tende a infinito (4x^3+2x^2+x+2)/(x^4-x^3+x^2+x+1) e, em seguida assinale a alternativa CORRETA:

1) 4
2) 0
3) -4
4) infinito
5) 4/3

Answer 1

Olá!


$\displaystyle\lim_{x\to\infty}\dfrac{4x^3+2x^2+x+2}{x^4-x^3+x^2+x+1}=\\\\\displaystyle\lim_{x\to\infty}\dfrac{x^3\left(4+\cancel{\dfrac{2}{x}}+\cancel{\dfrac{1}{x^2}}+\cancel{\dfrac{2}{x^3}}\right)}{x^4\left(1-\cancel{\dfrac{1}{x}}+\cancel{\dfrac{1}{x^2}}+\cancel{\dfrac{1}{x^3}}+\cancel{\dfrac{1}{x^4}}\right)}=\\\\\displaystyle\lim_{x\to\infty}\dfrac{x^3(4)}{x^4(1)}=\\\\\displaystyle\lim_{x\to\infty}\dfrac{4}{x}=0\\\\\\\boxed{\boxed{\displaystyle\lim_{x\to\infty}\dfrac{4x^3+2x^2+x+2}{x^4-x^3+x^2+x+1}=0}}$


Bons estudos!