What is the value of `a+b, if lim _(xto 0) (sin (ax) -ln (e ^(x)cos x))/(x sin (bx))=1/2` ?

Options

A. 1
B. 2
C. 3
D. `-1/2`

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{\sin(ax) - \ln(e^x \cos x)}{x \sin(bx)} = \frac{1}{2}, \]

we will follow these steps:

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0, leading to a \( \frac{0}{0} \) indeterminate form. Thus, we can apply L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
We differentiate the numerator and denominator separately.

**Numerator:**
\[ \frac{d}{dx}[\sin(ax) - \ln(e^x \cos x)] = a \cos(ax) - \frac{1}{e^x \cos x} \left(e^x \cos x - e^x (-\sin x)\right) \]
This simplifies to:
\[ a \cos(ax) - \frac{e^x \cos x + e^x \sin x}{e^x \cos x} = a \cos(ax) - \left(1 + \tan x\right) \]

**Denominator:**
\[ \frac{d}{dx}[x \sin(bx)] = \sin(bx) + bx \cos(bx) \]

### Step 3: Apply L'Hôpital's Rule
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{a \cos(ax) - (1 + \tan x)}{\sin(bx) + bx \cos(bx)} \]

### Step 4: Evaluate the limit as \( x \to 0 \)
Substituting \( x = 0 \):
- The numerator becomes \( a \cdot 1 - (1 + 0) = a - 1 \).
- The denominator becomes \( 0 + 0 = 0 \).

Since we still have a \( \frac{0}{0} \) form, we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
**Numerator:**
\[ \frac{d}{dx}[a \cos(ax) - (1 + \tan x)] = -a^2 \sin(ax) - \sec^2 x \]

**Denominator:**
\[ \frac{d}{dx}[\sin(bx) + bx \cos(bx)] = b \cos(bx) + \cos(bx) - b x \sin(bx) = (b + 1) \cos(bx) - b x \sin(bx) \]

### Step 6: Evaluate the limit again
Now we evaluate:
\[ \lim_{x \to 0} \frac{-a^2 \sin(ax) - \sec^2 x}{(b + 1) \cos(bx) - b x \sin(bx)} \]
Substituting \( x = 0 \):
- The numerator becomes \( 0 - 1 = -1 \).
- The denominator becomes \( (b + 1) \cdot 1 - 0 = b + 1 \).

Thus, we have:
\[ \frac{-1}{b + 1} = \frac{1}{2}. \]

### Step 7: Solve for \( b \)
Setting the equation:
\[ -1 = \frac{1}{2}(b + 1) \implies -2 = b + 1 \implies b = -3. \]

### Step 8: Find \( a \)
From the earlier limit evaluations, we found that \( a - 1 = 0 \implies a = 1 \).

### Step 9: Calculate \( a + b \)
Finally, we calculate:
\[ a + b = 1 - 3 = -2. \]

### Final Answer
Thus, the value of \( a + b \) is:
\[ \boxed{-2}. \]