Question
( begin{aligned}(n+2) !+(n+1) ! &=(n+1) ![n+2+1] &=(n+3) cdot(n+1) ! therefore lim _{n rightarrow infty} frac{(n+3) cdot(n+1) !}{(n+3)(n+2)(n+1) !} &=lim _{n rightarrow infty} frac{1}{n+2}=0 end{aligned} )

n +00 (n+2)! + (n+1)! Limit -, ne N= (n+3) (a) 0 (b) 1 (c)2 (d)-1
Solution
