Granica do policzenia :))

[aleksandrapyrpec]

Proszę o pomoc w obliczeniu granic:
\(\lim_{n\to \infty } \sqrt[n]{2n^3 - 3n^2 + 15}\)

\(\lim_{n\to \infty } ( \sqrt[3]{n^3 + 4n^2} - n )\)

\(\lim_{n\to \infty }( n \sqrt[3]{2} - \sqrt[3]{2n^3 + 5n - 7})\)

[irena]

\(\sqrt[3]{n^3+4n^2}-n=\frac{n^3+4n^2-n^3}{\sqrt[3]{n^6+8n^5+16n^4}+\sqrt[3]{n^6+4n^5}+\sqrt[3]{n^6}}=\\=\frac{4n^2}{n^2(\sqrt[3]{1+\frac{8}{n}+\frac{16}{n^2}}+\sqrt[3]{1+\frac{4}{n}}+1)}=\\=\frac{4}{\sqrt[3]{1+\frac{8}{n}+\frac{16}{n^2}}+\sqrt[3]{1+\frac{4}{n}}+1}\to\frac{4}{3}\)

[irena]

\(n\sqrt[3]{2}-\sqrt[3]{2n^3+5n-7}=n\sqrt[3]{2}-n\sqrt[3]{2+\frac{5}{n^2}-\frac{7}{n^3}}=\frac{2n^3-2n^3-5n+7}{n^2\sqrt[3]{4}+n^2\sqrt[3]{4+\frac{10}{n^2}-\frac{14}{n^3}}+n^2\sqrt[3]{(2+\frac{5}{n^2}-\frac{7}{n^3})^2}}=\\=\frac{-5n+7}{n^2(\sqrt[3]{4}+\sqrt[3]{4+\frac{10}{n^2}-\frac{7}{n^3}}+\sqrt[3]{(2+\frac{5}{n^2}-\frac{7}{n^3})^2})}=\\=\frac{-\frac{5}{n}-\frac{7}{n^2}}{\sqrt[3]{4}+\sqrt[3]{4+\frac{10}{n^2}-\frac{7}{n^3}}+\sqrt[3]{(2+\frac{5}{n^2}-\frac{7}{n^3})^2}}\to0\)

[irena]

\(\sqrt[n]{2n^3-3n^2+15}=\sqrt[n]{n^3}\cdot\sqrt[n]{2-\frac{3}{n}+\frac{15}{n^3}}=(\sqrt[n]{n})^3\cdot\sqrt[n]{2-\frac{3}{n}+\frac{15}{n^3}}\to1^3\cdot1=1\)