If `lim_(xrarr-1)(sin(x^3+bx^2+cx +d))/((sqrt(2+x)-1){log_e(x+2)}^2)` exists and is equal to l, then `b+d+l` is equal to

Options

A. 5
B. 6
C. 7
D. 4

Explanation

To solve the limit problem step by step, we will analyze the given expression and find the values of \( b \), \( d \), and \( l \).

### Step 1: Analyze the Limit
We start with the limit:
\[ \lim_{x \to -1} \frac{\sin(x^3 + bx^2 + cx + d)}{(\sqrt{2+x} - 1)(\log_e(x+2))^2} \]
For this limit to exist, both the numerator and denominator must approach 0 as \( x \to -1 \).

### Step 2: Evaluate the Denominator
First, we evaluate the denominator:
- As \( x \to -1 \), \( \sqrt{2+x} \to \sqrt{1} = 1 \), so \( \sqrt{2+x} - 1 \to 0 \).
- Also, \( \log_e(x+2) \to \log_e(1) = 0 \).

Thus, both parts of the denominator approach 0, confirming that we have a \( \frac{0}{0} \) indeterminate form.

### Step 3: Rationalize the Denominator
To simplify the limit, we can rationalize the denominator:
\[ \sqrt{2+x} - 1 = \frac{(2+x) - 1}{\sqrt{2+x} + 1} = \frac{x + 1}{\sqrt{2+x} + 1} \]
Now, we rewrite the limit:
\[ \lim_{x \to -1} \frac{\sin(x^3 + bx^2 + cx + d)}{\frac{x + 1}{\sqrt{2+x} + 1} (\log_e(x+2))^2} \]
This simplifies to:
\[ \lim_{x \to -1} \frac{\sin(x^3 + bx^2 + cx + d)(\sqrt{2+x} + 1)}{(x + 1)(\log_e(x+2))^2} \]

### Step 4: Determine the Behavior of the Numerator
For the limit to exist, the argument of the sine function must approach 0 as \( x \to -1 \):
\[ x^3 + bx^2 + cx + d \to 0 \]
Substituting \( x = -1 \):
\[ (-1)^3 + b(-1)^2 + c(-1) + d = -1 + b - c + d = 0 \]
Thus, we have:
\[ b - c + d = 1 \quad \text{(Equation 1)} \]

### Step 5: Analyze the Logarithm
Next, we analyze \( \log_e(x+2) \):
\[ \log_e(x+2) = \log_e(1) = 0 \text{ as } x \to -1 \]
Using the expansion \( \log_e(1 + u) \approx u \) for small \( u \):
\[ \log_e(x+2) \approx x + 1 \text{ as } x \to -1 \]
Thus, \( (\log_e(x+2))^2 \approx (x + 1)^2 \).

### Step 6: Substitute and Simplify
Substituting back into the limit:
\[ \lim_{x \to -1} \frac{\sin(x^3 + bx^2 + cx + d)(\sqrt{2+x} + 1)}{(x + 1)(x + 1)^2} \]
This simplifies to:
\[ \lim_{x \to -1} \frac{\sin(x^3 + bx^2 + cx + d)(\sqrt{2+x} + 1)}{(x + 1)^3} \]

### Step 7: Use the Sine Limit
Using the fact that \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \):
\[ \lim_{x \to -1} \frac{x^3 + bx^2 + cx + d}{(x + 1)^3} = \text{constant} \]
This means \( x^3 + bx^2 + cx + d \) must be of the form \( k(x + 1)^3 \).

### Step 8: Coefficients Comparison
We equate coefficients:
\[ x^3 + bx^2 + cx + d = k(x + 1)^3 = k(x^3 + 3x^2 + 3x + 1) \]
From this, we can equate coefficients:
- \( b = 3k \)
- \( c = 3k \)
- \( d = k \)

### Step 9: Solve for \( b, c, d \)
From Equation 1:
\[ 3k - 3k + k = 1 \implies k = 1 \]
Thus:
- \( b = 3 \)
- \( c = 3 \)
- \( d = 1 \)

### Step 10: Find \( l \)
Now, substituting back, we find \( l \):
\[ l = 2 \quad \text{(as derived from the limit)} \]

### Final Calculation
Now we calculate \( b + d + l \):
\[ b + d + l = 3 + 1 + 2 = 6 \]

### Answer
Thus, the final answer is:
\[ \boxed{6} \]