Răspuns:
Explicație pas cu pas:
[tex]\lim_{x \to +\infty} (\sqrt[3]{x+1}-\sqrt[3]{x})=( \infty-\infty)= \lim_{x \to +\infty} \frac{ (\sqrt[3]{x+1}-\sqrt[3]{x})(\sqrt[3]{(x+1)^{2}} +\sqrt[3]{x(x+1)}+\sqrt[3]{x^{2}}) }{\sqrt[3]{(x+1)^{2}} +\sqrt[3]{x(x+1)}+\sqrt[3]{x^{2}} } = \lim_{x \to +\infty} \frac{(\sqrt[3]{x+1})^{3}-(\sqrt[3]{x})^{3} }{\sqrt[3]{(x+1)^{2}} +\sqrt[3]{x(x+1)}+\sqrt[3]{x^{2}} } = \lim_{x \to +\infty} \frac{x+1-x }{\sqrt[3]{(x+1)^{2}} +\sqrt[3]{x(x+1)}+\sqrt[3]{x^{2}} } =[/tex]
[tex]=\lim_{x \to +\infty} \frac{1}{\sqrt[3]{(x+1)^{2}} +\sqrt[3]{x(x+1)}+\sqrt[3]{x^{2}} } =\frac{1}{\infty}=0[/tex]