If `lim_(xto0) (sin 2x+asinx)/x^3=b` then the value of b-2a is equal to

Answer

To solve the limit problem given, we need to find the value of \( b - 2a \) where

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

### Step 1: Identify the form of the limit
As \( x \to 0 \), both \( \sin(2x) \) and \( \sin(x) \) approach 0. Thus, the limit takes the form \( \frac{0 + 0}{0} \), which is an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit is of the form \( \frac{0}{0} \), we can 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: Evaluate the limit again
Substituting \( x = 0 \) into the new limit gives:

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

This is still an indeterminate form \( \frac{C}{0} \) (where \( C = 2 + a \)). For the limit to exist, the numerator must also approach 0 as \( x \to 0 \):

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

### Step 4: Substitute \( a \) back into the limit
Now substituting \( a = -2 \) back into the limit, we have:

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

### Step 5: Apply L'Hôpital's Rule again
This limit is still \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:

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

### Step 6: Evaluate the limit again
Substituting \( x = 0 \):

\[ \frac{2\cos(0) - 2\cos(0)}{3(0)^2} = \frac{2 - 2}{0} = \frac{0}{0}. \]

### Step 7: Apply L'Hôpital's Rule a third time
We apply L'Hôpital's Rule again:

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

### Step 8: Evaluate the limit again
Substituting \( x = 0 \):

\[ \frac{-4\sin(0) + 2\sin(0)}{6(0)} = \frac{0}{0}. \]

### Step 9: Apply L'Hôpital's Rule a fourth time
Applying L'Hôpital's Rule one more time:

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

### Step 10: Evaluate the limit
Substituting \( x = 0 \):

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

### Step 11: Find \( b \)
Thus, we have \( b = -1 \).

### Step 12: Calculate \( b - 2a \)
Now substituting \( a = -2 \):

\[ b - 2a = -1 - 2(-2) = -1 + 4 = 3. \]

### Final Answer
The value of \( b - 2a \) is \( 3 \).