Question

${lim}_{x\to \infty }{\left(\frac{n!}{(n+1)!-n!}\right)}^{}$ is equal to

• A. $0$
• B. $\infty$
• C. $1$
• D. none of these

Answer 1

The correct option is A $0$

As $(n+1)!=(n+1)\times n!$

$\Rightarrow \underset{n\to \infty }{lim}\frac{n!}{(n+1)!-n!}$

$=\frac{n!}{(n+1)n!-n!}$

$=\frac{n!}{n!(n+1-1)}$

$=\underset{n\to \infty }{lim}\frac{1}{n}=0$

$\Rightarrow$ Option (A).