Question

Determine o limite e depois use a regra de Lá hopital

Answer 1

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O limite vale 2.

$\lim_{n \to -\infty} \dfrac{4n^2-n}{2n^2-5} = \\\\\\ \lim_{n \to -\infty} \dfrac{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}-\dfrac{5}{n^2}} = \\\\\\ \lim_{n \to -\infty} \dfrac{4-{\dfrac{1}{n}}}{2-\dfrac{5}{n^2}} = \dfrac{4}{2} = 2$