Question

Calcule, caso existam, os limites indicados abaixo

Answer 1

Resposta:

a) 0.

b) 4/3.

c) Limite inexistente.

Explicação passo a passo:

a)

$\lim_{x \to \infty} \frac{2x^{3}+4x^{2}-1 }{3x^{4}+2x-2} = \lim_{x \to \infty} \frac{6x^{2}+8x}{12x^{3}+2} = \lim_{x \to \infty} \frac{12x+8}{36x^{2}} = \lim_{x \to \infty} \frac{12}{72x} = 0.$

b)

$\lim_{x \to \infty} \frac{4x^{4}+x+3 }{3x^{4}+x^{3}-1} = \lim_{x \to \infty} \frac{16x^{3}+1}{12x^{3}+3x^{2}} = \lim_{x \to \infty} \frac{48x^{2}}{36x^{2}+6x} = \lim_{x \to \infty} \frac{96x}{72x+6} = \lim_{x \to \infty} \frac{96}{72} = \frac{96}{72} = \frac{4}{3}$

c)

$\lim_{x \to \infty} \frac{4x^{3}+2x^{2}-x+3 }{2x^{2} +3x-8} = \lim_{x \to \infty} \frac{12x^{2}+4x-1}{4x+3} = \lim_{x \to \infty} \frac{24x+4}{4} =$

= +∞.