Question

lim ((n+2)!+(n+1)!)/(n+3)!
n-бескон.

Answer 1

Lim ((n+2)!+(n+1)!)/(n+3)! = lim ((n+1)!*(n+2) +(n+1)!)/((n+1)!*(n+2)*(n+3))=
lim((n+1)!(n+2+1))/((n+1)!*(n+2)*(n+3))= lim (n+3)/ ((n+2)*(n+3))=
lim (n+3)/(n²+5n+6)=lim (1/n+3/n²)/(1+5/n+6/n²)= 0/1=0