Question

Salut, ma puteti ajuta la problema numarul 488...?

Answer 1

$\displaystyle Pe~intervalul~(n,n+1),~n>1,~avem~x^3+x+1>x^3>x^2. \\ \\ \sqrt{x^3+x+1}> \sqrt{x^2}=x \\ \\ Avem~0<\int\limits_n^{n+1} \frac{1}{\sqrt{x^3+x+1}} \mathrm{d}x< \int\limits_n^{n+1} \frac{1}{x} \mathrm{d}x= \ln \frac{n+1}{n} \to 0. \\ \\ Deci~limita~este~0. \\ \\ (Am~folosit~criteriul~clestelui.)$