Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $1/2$

Answer 1

The correct option is A $1$

Now,

$\underset{n\to \infty }{lim}\frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}$

$=\underset{n\to \infty }{lim}\frac{(n+1)!\left\{\right(n+2)+1\}}{(n+1)!\left\{\right(n+2)-1\}}$

$=\underset{n\to \infty }{lim}\frac{\left\{\right(n+3\left)\right\}}{\left\{\right(n+1\left)\right\}}$

$=\underset{n\to \infty }{lim}\frac{\left\{\right(1+\frac{3}{n}\left)\right\}}{\left\{\right(1+\frac{1}{n}\left)\right\}}$

$=\frac{1}{1}$ [ Using the property $\underset{n\to \infty }{lim}\frac{1}{n}=0$]

$=1$.