Test Index

MHTCET 13 Apr 2026 Shift 2 Paper

© examsnet.com

Question : 111 of 150

Marks: +1, -0

The value of

\lim _{n \rightarrow \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}

\_\_\_\_

-1
0
1
2

Solution:

Concept:

Simplify factorials using

(n+2)! = (n+2)(n+1)!

, then evaluate the limit.

Explanation:

Rewrite

(n+2)!

as

(n+2)(n+1)!

.

Substitute into the given expression:

\frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!} = \frac{(n+2)(n+1)!+(n+1)!}{(n+2)(n+1)!-(n+1)!}

Factor out

(n+1)!

from numerator and denominator:

\frac{(n+1)![(n+2)+1]}{(n+1)![(n+2)-1]}

Cancel

(n+1)!

to get:

\frac{n+3}{n+1}

Now take the limit as

n \to \infty

:

\lim_{n \to \infty} \frac{n+3}{n+1} = \lim_{n \to \infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}} = 1

Answer:

1

Correct option: C.

© examsnet.com

Go to Question:

Prev Question

Next Question