Question

P$\lim_{x \to +\infty} \sqrt[3]{ x^{3} + x^{2} }-x$or favor me ajudem com esse limite.

Answer 1

Boa noite!

$\displaystyle{\lim_{x\to + \infty}{\sqrt[3]{x^3+x^2}-x}}\\ \displaystyle{\lim_{x\to + \infty}{\sqrt[3]{x^3\left(1+\frac{1}{x}\right)}-x}}\\ \displaystyle{\lim_{x\to + \infty}{x\sqrt[3]{1+\frac{1}{x}}-x}}\\ \displaystyle{\lim_{x\to + \infty}{x\left(\sqrt[3]{1+\frac{1}{x}}-1\right)}}\\ \displaystyle{\lim_{x\to + \infty}\frac{\sqrt[3]{1+\frac{1}{x}}-1}{\frac{1}{x}}\text{ L'Hospital}}\\ \displaystyle{\lim_{x\to + \infty}\frac{\frac{1}{3}\left(1+\frac{1}{x}\right)^{-\frac{2}{3}}\cdot\left(1+\frac{1}{x}\right)'}{\frac{-1}{x^2}}}\\ \displaystyle{\lim_{x\to + \infty}\frac{\frac{1}{3}\left(1+\frac{1}{x}\right)^{-\frac{2}{3}}\cdot\frac{-1}{x^2}}{\frac{-1}{x^2}}}\\ \displaystyle{\lim_{x\to + \infty}\frac{1}{3}\left(1+\frac{1}{x}\right)^{-\frac{2}{3}}=\frac{1}{3}\left(1+\frac{1}{\infty}\right)^{-\frac{2}{3}}}\\ \displaystyle{\frac{1}{3}}$

Espero ter ajudado!