Question

помогите пожалуйста​

Answer 1

Ответ:

N258

0

N259

1

Пошаговое объяснение:

N258

$\lim_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+3)!} =\lim_{n \to \infty} \frac{(n+1)!*(n+2+1)}{(n+1)!*(n+2)*(n+3)} =\\=\lim_{n \to \infty} \frac{n+3}{n^{2}+5n+6} =\lim_{n \to \infty} \frac{\frac{n}{n^{2}}+\frac{3}{n^{2}}}{1+\frac{5n}{n^{2}}+\frac{6}{n^{2}} } =\\=\lim_{n \to \infty} \frac{\frac{1}{n}+\frac{3}{n^{2}}}{1+\frac{5}{n}+\frac{6}{n^{2}} }=\frac{0+0}{1+0+0}=0$

N259

$\lim_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!} =\\=\lim_{n \to \infty} \frac{(n+1)!*(n+2+1)}{(n+1)!*(n+2-1)} =\\=\lim_{n \to \infty} \frac{n+3}{n+1} =\lim_{n \to \infty} \frac{1+\frac{3}{n} }{1+\frac{1}{n}} =\\=\frac{1+0}{1+0} =1$