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a fost răspuns

[tex]\lim _{x\to \infty }\:\left(\sqrt[3]{x+1}-\sqrt[3]{x}\right)\\\\\lim _{x\to \infty }\frac{\left(\sqrt[3]{X+1}-\sqrt[3]{X}\right)^2}{\left(\sqrt[3]{X+1}+\sqrt[3]{X}\right)}[/tex]pornind de la forma fara fractie am aplificat cu conjugata si am ajuns la forma cu fractie ,nu stiu cum sa mai continui .

Răspuns :

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]