Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$ Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot 2 \ldots(n-1)n}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n}=\lim_{n \to \infty}\frac{(1 \cdot 2 \ldots(n-1) n)\cdot((n+1)(n+2)(n+3)-1)}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n} =\lim_{n \to \infty} \frac{(n+1)(n+2)(n+3)-1}{(n+1)(n+2)\cdot n} = \frac{1}{1}=1$$ 
The answer is 1 because we've got $n^3$ in numerator and $n^3 $ in denominator, and the constants are both equals 1 so the limit can be calculated as $\frac{1}{1}$. (Sorry I don't remember that rule. Could you remain me please?)

You can more easily write the expression (cancelling as much as possible) as $$\frac {n+3}n-\frac 1{n(n+1)(n+2)}=1+\frac 3n-\frac 1{n(n+1)(n+2)}$$
Then the limit is very easy to see.
