`lim_(xrarr-7) ([x]^(2)+15[x]+56)/(sin(x+7)sin(x+8))=` (where [.] denotes the greatest integer function)

Options

A. is 0
B. is 1
C. is `-1`
D. does not exist

Explanation

To solve the limit

\[ \lim_{x \to -7} \frac{[x]^2 + 15[x] + 56}{\sin(x + 7) \sin(x + 8)} \]

where \([x]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Determine the value of \([x]\) as \(x\) approaches \(-7\)

As \(x\) approaches \(-7\), the greatest integer function \([x]\) will equal \(-8\) for values of \(x\) slightly less than \(-7\) (i.e., \(x \to -7^-\)), and it will equal \(-7\) for values of \(x\) slightly greater than \(-7\) (i.e., \(x \to -7^+\)).

### Step 2: Evaluate the numerator

For \(x \to -7^-\):
\[ [x] = -8 \implies [x]^2 + 15[x] + 56 = (-8)^2 + 15(-8) + 56 = 64 - 120 + 56 = 0 \]

For \(x \to -7^+\):
\[ [x] = -7 \implies [x]^2 + 15[x] + 56 = (-7)^2 + 15(-7) + 56 = 49 - 105 + 56 = 0 \]

### Step 3: Evaluate the denominator

Now, we evaluate the denominator \(\sin(x + 7) \sin(x + 8)\):

As \(x \to -7\):
- For \(x \to -7^-\):
\[ \sin(x + 7) = \sin(0) = 0 \quad \text{and} \quad \sin(x + 8) = \sin(1) \neq 0 \]

- For \(x \to -7^+\):
\[ \sin(x + 7) = \sin(0) = 0 \quad \text{and} \quad \sin(x + 8) = \sin(1) \neq 0 \]

### Step 4: Apply L'Hôpital's Rule

Since both the numerator and denominator approach \(0\) as \(x \to -7\), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

1. Differentiate the numerator:
\[ \frac{d}{dx}([x]^2 + 15[x] + 56) = 0 \quad \text{(since \([x]\) is constant in intervals)} \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(\sin(x + 7) \sin(x + 8)) = \cos(x + 7) \sin(x + 8) + \sin(x + 7) \cos(x + 8) \]

### Step 5: Evaluate the limit again

Now we evaluate the limit again using L'Hôpital's Rule:

For \(x \to -7\):
- The numerator is \(0\).
- The denominator becomes:
\[ \cos(-7 + 7) \sin(-7 + 8) + \sin(-7 + 7) \cos(-7 + 8) = \cos(0) \sin(1) + 0 = 1 \cdot \sin(1) = \sin(1) \neq 0 \]

Thus, we have:
\[ \lim_{x \to -7} \frac{0}{\sin(1)} = 0 \]

### Conclusion

The limit is:

\[ \lim_{x \to -7} \frac{[x]^2 + 15[x] + 56}{\sin(x + 7) \sin(x + 8)} = 0 \]