Question

Найти предел на фото

Answer 1

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