If `lim_(x to 0) ("sin" 2x + a "sin" x)/(x^(3)) =b` (finite), then `(ab)^(2)` equals…..

Answer

To solve the limit problem, we need to find the values of \( a \) and \( b \) such that

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

is finite. Let's go through the steps systematically.

### Step 1: Identify the form of the limit
When we substitute \( x = 0 \) into the expression, we get:

\[ \frac{\sin 2(0) + a \sin(0)}{0^3} = \frac{0 + 0}{0} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

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

### Step 3: Substitute \( x = 0 \) again
Now, substituting \( x = 0 \) gives:

\[ \frac{2 \cos 0 + a \cos 0}{3 \cdot 0^2} = \frac{2 + a}{0} \]

This indicates that for the limit to be finite, the numerator must equal zero:

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

### Step 4: Substitute \( a \) back into the limit
Now, substitute \( a = -2 \) into the limit expression:

\[ \lim_{x \to 0} \frac{\sin 2x - 2 \sin x}{x^3} \]

### Step 5: Apply L'Hôpital's Rule again
This again gives the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:

\[ \lim_{x \to 0} \frac{2 \cos 2x - 2 \cos x}{3x^2} \]

### Step 6: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

\[ \frac{2 \cos 0 - 2 \cos 0}{3 \cdot 0^2} = \frac{2 - 2}{0} = \frac{0}{0} \]

We apply L'Hôpital's Rule again:

\[ \lim_{x \to 0} \frac{-4 \sin 2x + 2 \sin x}{6x} \]

### Step 7: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

\[ \frac{-4 \sin 0 + 2 \sin 0}{6 \cdot 0} = \frac{0}{0} \]

We apply L'Hôpital's Rule once more:

\[ \lim_{x \to 0} \frac{-8 \cos 2x + 2 \cos x}{6} \]

### Step 8: Substitute \( x = 0 \) one last time
Substituting \( x = 0 \):

\[ \frac{-8 \cos 0 + 2 \cos 0}{6} = \frac{-8 + 2}{6} = \frac{-6}{6} = -1 \]

Thus, we find that \( b = -1 \).

### Step 9: Calculate \( (ab)^2 \)
Now, we have \( a = -2 \) and \( b = -1 \):

\[ ab = (-2)(-1) = 2 \]

Thus,

\[ (ab)^2 = 2^2 = 4 \]

### Final Answer
Therefore, the value of \( (ab)^2 \) is

\[ \boxed{4} \]