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?)

If $m=n$, and $a_n,b_m\neq 0$, then you can say the following:$$\lim_{x \to \infty}\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{a_n}{b_m}.$$
To prove this, you can factor out the $x$ terms from the numerator and denominator like so: $$\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{x^n}{x^m}\cdot\frac{a_n+\cdots+\frac{a_0}{x^n}}{b_m+\cdots+\frac{b_0}{x^m}}.$$
Now since $n = m$, $x^n/x^m=1$, the limit becomes $$\lim_{x \to \infty}\frac{a_n+\cdots+\frac{a_0}{x^n}}{b_m+\cdots+\frac{b_0}{x^m}} =\frac{\lim_{x \to \infty}a_n+\cdots+\frac{a_0}{x^n}}{\lim_{x \to \infty}b_m+\cdots+\frac{b_0}{x^m}}=\frac{a_n}{b_m}.$$
since all the terms with an $x$ in the denominator go to $0$ as $x \to \infty$. You can also use a similar technique to show that if $m > n$ then the limit is $0$.
