Q56 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
The value of $\lim_{n\to\infty} \dfrac{(n+2)! + (n+1)!}{(n+2)! - (n+1)!}$ is ____
- A$-1$
- B$0$
- Correct answerC$1$
- D$2$
Solution
The given limit is $\lim_{n\to\infty} \dfrac{(n+2)! + (n+1)!}{(n+2)! - (n+1)!}$
Taking $(n+1)!$ common from both numerator and denominator:
$\lim_{n\to\infty} \dfrac{(n+1)![(n+2) + 1]}{(n+1)![(n+2) - 1]}$
Cancelling $(n+1)!$ from numerator and denominator:
$\lim_{n\to\infty} \dfrac{n+3}{n+1}$
Dividing numerator and denominator by $n$:
$\lim_{n\to\infty} \dfrac{1 + \dfrac{3}{n}}{1 + \dfrac{1}{n}}$
As $n \to \infty$, $\dfrac{1}{n} \to 0$. Substituting the limit:
$\dfrac{1 + 0}{1 + 0} = 1$
Answer: $1$
Taking $(n+1)!$ common from both numerator and denominator:
$\lim_{n\to\infty} \dfrac{(n+1)![(n+2) + 1]}{(n+1)![(n+2) - 1]}$
Cancelling $(n+1)!$ from numerator and denominator:
$\lim_{n\to\infty} \dfrac{n+3}{n+1}$
Dividing numerator and denominator by $n$:
$\lim_{n\to\infty} \dfrac{1 + \dfrac{3}{n}}{1 + \dfrac{1}{n}}$
As $n \to \infty$, $\dfrac{1}{n} \to 0$. Substituting the limit:
$\dfrac{1 + 0}{1 + 0} = 1$
Answer: $1$
Hence, the correct answer is option: 'C'.