The value of `lim_(x rarr oo) ((x+2)!+ (x+1)!)/((x+2)!-(x+1)!)` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the limit problem \( \lim_{x \to \infty} \frac{(x+2)! + (x+1)!}{(x+2)! - (x+1)!} \), we can follow these steps:

### Step 1: Rewrite the factorials
We can express \( (x+2)! \) in terms of \( (x+1)! \):
\[ (x+2)! = (x+2)(x+1)! \]
Substituting this into the limit expression gives:
\[ \lim_{x \to \infty} \frac{(x+2)(x+1)! + (x+1)!}{(x+2)(x+1)! - (x+1)!} \]

### Step 2: Factor out \( (x+1)! \)
Now we can factor \( (x+1)! \) out of both the numerator and the denominator:
\[ = \lim_{x \to \infty} \frac{(x+1)! \left((x+2) + 1\right)}{(x+1)! \left((x+2) - 1\right)} \]
This simplifies to:
\[ = \lim_{x \to \infty} \frac{(x+1)! (x+3)}{(x+1)! (x+1)} \]

### Step 3: Cancel \( (x+1)! \)
Since \( (x+1)! \) is common in both the numerator and the denominator, we can cancel it:
\[ = \lim_{x \to \infty} \frac{x+3}{x+1} \]

### Step 4: Simplify the limit
Now we can simplify the expression:
\[ = \lim_{x \to \infty} \frac{x+3}{x+1} = \lim_{x \to \infty} \frac{1 + \frac{3}{x}}{1 + \frac{1}{x}} \]

### Step 5: Evaluate the limit
As \( x \to \infty \), \( \frac{3}{x} \to 0 \) and \( \frac{1}{x} \to 0 \):
\[ = \frac{1 + 0}{1 + 0} = 1 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{1} \]