$\underset{n→∞}{\lim}\frac{(n+2)!+(n+3)!}{(n+1)!-(n+3)!}$ is equal to

-1

$\underset{n→∞}{\lim}\frac{(n+2)!+(n+3)!}{(n+1)!-(n+3)!}=\underset{n→∞}{\lim}\frac{(n+2)!1+n+3]}{(n+1)![1-(n+3)(n+2)]}$

$=\underset{n→∞}{\lim}\frac{(n+1)!(n+2)(n+4)}{(n+1)![1-(n^2+5n+6)]}=\underset{n→∞}{\lim}\frac{n^2+6n+8}{-n^2-5n-5}=\underset{n→∞}{\lim}\frac{1+\frac{6}{n}+\frac{8}{n^2}}{-1-\frac{5}{n}-\frac{5}{n^2}}=-1$