Question

Calcule, caso exista, limite de n maior que o infinito an: an = n³ + 3n + 1/ 4n³ + 2

Answer 1

Resposta:

Resolvamos o limite dado pela Regra de l'Hôpital:

$\lim_{n \to \infty} \frac{n^3+3n+1}{4n^3+2}\\\\= \lim_{n \to \infty} \frac{3n^2+3}{12n^2}\\\\= \lim_{n \to \infty} \frac{6n}{24n}\\\\= \lim_{n \to \infty} \frac{6}{24}\\\\= \frac{6}{24} = \boxed{\frac{1}{4}}$