Question

помогите с пределом пожалуйста

Answer 1

$\lim\limits _{n \to \infty}(\sqrt{\frac{n^4+n^3}{n^2+1}}-\sqrt {n^2-1})=\lim\limits _{n \to \infty}\frac{\sqrt{n^4+n^3}-\sqrt{(n^2-1)(n^2+1)}}{\sqrt{n^2+1}}=\\\\=\lim\limits _{n \to \infty}\frac{\sqrt{n^3(n+1)}-\sqrt{(n+1)(n-1)(n^2+1)}}{\sqrt{n^2+1}}=\\\\=\lim\limits _{n \to \infty}\frac{\sqrt{n+1}\cdot (\sqrt{n^3}-\sqrt{(n-1)(n^2+1)})}{\sqrt{n^2+1}}=$

$=\lim\limits _{n \to \infty}\frac{\sqrt{n+1}\cdot (n^3-(n-1)(n^2+1))}{\sqrt{n^2+1}\cdot (\sqrt{n^3}+\sqrt{(n-1)(n^2+1)}}=\lim\limits _{n \to \infty}\frac{\sqrt{n+1}\cdot (n^2-n+1)}{\sqrt{n^2+1}\cdot (\sqrt{n^3}+\sqrt{n^3+n-n^2-1})}=\\\\=\lim\limits _{n \to \infty}\frac{\sqrt{n}\cdot n^2}{\sqrt{n^2}\cdot 2\sqrt{n^3}}=\lim\limits _{n \to \infty}\frac{n^{5/2}}{2n^{5/2}}=\frac{1}{2}$