limites

resposta=0

Temos:

[latex]y = \lim_{x \rightarrow +\infty}\left [\sqrt[3]{x+1}-\sqrt[3]{x} \right ][/latex]

Com isso,

[latex]y = \lim_{x \rightarrow +\infty}\left [\sqrt[3]{x} \cdot \left (\sqrt[3]{1+\frac{1}{x}}-1 \right ) \right ][/latex]

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\sqrt[3]{1+\frac{1}{x}}-1}{\frac{1}{\sqrt[3]{x}}} \right ][/latex]

Pela regra de L'Hôpital,

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\frac{1}{3}\cdot \frac{\sqrt[3]{1+\frac{1}{x}}}{1 + \frac{1}{x}}\cdot \left ( -\frac{1}{x^2} \right )}{-\frac{1}{3}\cdot \frac{1}{x\cdot \sqrt[3]{x}}} \right ][/latex]

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\frac{\sqrt[3]{1+\frac{1}{x}}}{1 + \frac{1}{x}}\cdot \frac{1}{x^2}}{\frac{1}{x\cdot \sqrt[3]{x}}} \right ][/latex]

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\sqrt[3]{1+\frac{1}{x}}}{1 + \frac{1}{x}}\cdot \frac{x\cdot \sqrt[3]{x}}{x^2}\right ][/latex]

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\sqrt[3]{1+\frac{1}{x}}}{1 + \frac{1}{x}}\cdot \frac{\sqrt[3]{x}}{x}\right ][/latex]

[latex]y = \lim_{x \rightarrow +\infty}\left [ \frac{\sqrt[3]{x+1}}{x + 1}\right ][/latex]

Fazendo k^3 = x + 1, temos:

[latex]y = \lim_{k \rightarrow +\infty}\left [ \frac{k}{k^3}\right ] = \lim_{k \rightarrow +\infty}\left [ \frac{1}{k^2}\right ][/latex]


[latex]y = \lim_{k \rightarrow +\infty}\left [ \frac{1}{k^2}\right ] = 0[/latex]



[latex]y = 0[/latex]