Znajdz granice..

[szczud1]

Witam! Mam prośbę, czy mogłby ktos krok po kroku pokazac jak obliczac takie granice ?

\(\displaystyle{ \lim_{x\to\infty}}\) \(\displaystyle{ \frac{1}{ n^{3} }}\) \(\displaystyle{ {n+1\choose 2}}\) \(\displaystyle{ \sin (n!+n^{2})}\)



\(\displaystyle{ \lim_{x\to\infty}}\) (\(\displaystyle{ \sqrt[3]{1- \frac{1}{n} }}\) -1 ) n


Z góry dziekuje

[sir_matin]

\(\displaystyle{ \lim_{n\to\infty}( \sqrt[3]{1- \frac{1}{n} } -1)n=\lim_{n\to\infty}( \sqrt[3]{1- \frac{1}{n} } -1)n \frac{(\sqrt[3]{1- \frac{1}{n} } )^{2}+\sqrt[3]{1- \frac{1}{n} } +1}{(\sqrt[3]{1- \frac{1}{n} } )^{2}+\sqrt[3]{1- \frac{1}{n} } +1} = \\ = \lim_{n\to\infty} - \frac{1}{(\sqrt[3]{1- \frac{1}{n} } )^{2}+\sqrt[3]{1- \frac{1}{n} } +1} =- \frac{1}{3}}\)

[szczud1]

pomyslowo , nie wpadl bym na to.. a jak z ta pierwsza?

[Szemek]

\(\displaystyle{ \lim_{n\to\infty} \frac{1}{ n^{3} } {n+1\choose 2} \sin (n!+n^{2}) = \lim_{n\to\infty} \frac{n(n+1)\sin (n!+n^2)}{2n^3} = 0 \\
0 -\frac{n(n+1)}{2n^3} \frac{n(n+1)\sin (n!+n^2)}{2n^3} \frac{n(n+1)}{2n^3} \to 0}\)