Question

Найти предел на фото

Answer 1

$\lim\limits _{n \to +\infty}\frac{ln\, n+ln(n+1)}{\sqrt{n}}=\lim\limits _{n \to +\infty}\frac{ln(n^2+n)}{\sqrt{n}}= \lim_{n \to \infty}\frac{(ln(n^2+n))'}{(\sqrt{n})'}=\\\\=\lim\limits _{n \to +\infty}\frac{\frac{2n+1}{n^2+n}}{\frac{1}{2\sqrt{n}}}=\lim\limits _{n \to +\infty}\frac{(2n+1)\cdot 2\sqrt{n}}{n^2+n}}=\lim\limits _{n \to +\infty}\frac{4n^{3/2}+2n^{1/2}}{n^2+n}=\\\\=\lim\limits _{n \to +\infty}\frac{4n^{3/2}}{n^2}=\lim\limits _{n \to \infty}\frac{4}{n^{1/2}}=\Big [\; \frac{4}{+\infty }=0\; \Big ]=0$