Question

Calcule o limite a baixo:
lim x->-2. x^3 - 5x -2/ x^4 + 4x^3 + x^2 - 5x + 2

Answer 1

$\mathrm{\lim_{x\to a}\dfrac{M(x)}{N(x)}=\lim_{x\to-2}\dfrac{x^3-5x-2}{x^4+4x^3+x^2-5x+2}}\\\\\\ \mathbf{-2\ \'e\ raiz\ de\ M(x),\ logo:}\\\\ \mathrm{x^3-5x-2=\big(x-(-2)\big)\bigg(\dfrac{x^3-5x-2}{\big(x-(-2)\big)}\bigg)=}\\\\ \mathrm{=(x+2)\bigg(\dfrac{x^3-5x-2}{(x+2)}\bigg)=(x+2)(x^2-2x-1)}\\\\\\ \mathbf{-2\ \'e\ raiz\ de\ N(x),\ logo:}\\\\ \mathrm{x^4+4x^3+x^2-5x+2=}\\\\ \mathrm{(x+2)\bigg(\dfrac{x^4+4x^3+x^2-5x+2}{(x+2)}\bigg)=}\\\\ \mathrm{=(x+2)(x^3+2x^2-3x+1)}$

$\mathbf{Portanto,\ teremos\ que:}\\\\\ \mathrm{\lim_{x\to -2}\dfrac{(x+2)(x^2-2x-1)}{(x+2)(x^3+2x^2-3x+1)}=}\\\\ \mathrm{=\lim_{x\to -2}\dfrac{x^2-2x-1}{x^3+2x^2-3x+1}=\dfrac{(-2)^2-2(-2)-1}{(-2)^3+2(-2)^2-3(-2)+1}=}\\\\ \mathrm=\dfrac{4+4-1}{-8+8+6+1}=\dfrac{7}{7}=\mathbf{1}}$