Question

The value of $\underset{n\to \infty }{\lim }\frac{\left(n+2\right)!+\left(n+1\right)!}{\left(n+2\right)!-\left(n+1\right)!}$ is

(a) 0
(b) −1
(c) 1
(d) none of these

Answer 1

(c) 1
$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{\left(n+2\right)!+\left(n+1\right)!}{\left(n+2\right)!-\left(n+1\right)!} \\ \mathrm{Dividing}{\mathrm{N}}^r\&{\mathrm{D}}^r\mathrm{by}\left(n+1\right)!: \\ \Rightarrow \underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{\left(n+2\right)\left(n+1\right)!}{\left(n+1\right)!}}+1}{{\displaystyle \frac{\left(n+2\right)\left(n+1\right)!}{\left(n+1\right)!}}-1} \\ =\underset{n\to \infty }{\lim }\frac{n+2+1}{n+2-1} \\ =\underset{n\to \infty }{\lim }\frac{n+3}{n+1} \\ =\underset{n\to \infty }{\lim }\frac{1+{\displaystyle \frac{3}{n}}}{1+{\displaystyle \frac{1}{n}}} \\ =1 \end{gathered}$$