If `lim_(xrarroo)(ae^(x)+b cos x+c +dx)/(xsin^(2)x)=3,` then the value of `272(abd)/(c^(3))` is equal to

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{ae^x + b \cos x + c + dx}{x \sin^2 x} = 3, \]

we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \(x = 0\) into the expression:

\[ ae^0 + b\cos(0) + c + d(0) = a + b + c. \]

The denominator becomes:

\[ 0 \cdot \sin^2(0) = 0. \]

Thus, we have the indeterminate form \( \frac{0}{0} \), which means we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(ae^x + b\cos x + c + dx) = ae^x - b\sin x + d. \]

**Denominator:**
\[ \frac{d}{dx}(x \sin^2 x) = \sin^2 x + x \cdot 2\sin x \cos x = \sin^2 x + 2x \sin x \cos x. \]

### Step 3: Evaluate the limit again
Now we evaluate the limit again:

\[ \lim_{x \to 0} \frac{ae^x - b\sin x + d}{\sin^2 x + 2x \sin x \cos x}. \]

Substituting \(x = 0\):

**Numerator:**
\[ ae^0 - b\sin(0) + d = a + d. \]

**Denominator:**
\[ \sin^2(0) + 2(0)\sin(0)\cos(0) = 0. \]

Again, we have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again
We differentiate the numerator and denominator again.

**Numerator:**
\[ \frac{d}{dx}(ae^x - b\sin x + d) = ae^x - b\cos x. \]

**Denominator:**
\[ \frac{d}{dx}(\sin^2 x + 2x \sin x \cos x) = 2\sin x \cos x + 2(\sin x \cos x + x(\cos^2 x - \sin^2 x)). \]

### Step 5: Evaluate the limit again
Now we evaluate the limit again:

\[ \lim_{x \to 0} \frac{ae^x - b\cos x}{2\sin x \cos x + 2(\sin x \cos x + x(\cos^2 x - \sin^2 x))}. \]

Substituting \(x = 0\):

**Numerator:**
\[ a - b. \]

**Denominator:**
\[ 2(0)(1) + 2(0)(1) = 0. \]

Again, we have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 6: Solve the equations
From the previous evaluations, we have:

1. \( a + b + c = 0 \)
2. \( a + d = 0 \)
3. \( a - b = 0 \)

From \( a - b = 0 \), we have \( a = b \).

Substituting \( b = a \) into \( a + b + c = 0 \):

\[ a + a + c = 0 \implies 2a + c = 0 \implies c = -2a. \]

Substituting \( d = -a \) into the limit equation gives:

\[ \lim_{x \to 0} \frac{ae^x + ae^0 - 2a - a}{x \sin^2 x} = 3. \]

### Step 7: Solve for \(a\)
We need to find \(a\) such that the limit equals 3. After applying L'Hôpital's Rule sufficiently, we find that:

\[ \frac{3a}{0} = 3 \implies a = 9. \]

### Step 8: Find values of \(a\), \(b\), \(c\), and \(d\)
Thus, we have:

- \( a = 9 \)
- \( b = 9 \)
- \( c = -18 \)
- \( d = -9 \)

### Step 9: Calculate \( \frac{272(abd)}{c^3} \)
Now we can substitute these values into the expression:

\[ \frac{272(9 \cdot 9 \cdot -9)}{(-18)^3} = \frac{272(-729)}{-5832}. \]

Calculating this gives:

\[ \frac{272 \cdot 729}{5832} = \frac{198144}{5832} = 34. \]

Thus, the final answer is:

\[ \boxed{34}. \]