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 form
As \(x\) approaches \(0\), both the numerator and denominator approach \(0\). This gives us a \(0/0\) indeterminate form, which allows us to apply L'Hôpital's rule.

### Step 2: Differentiate the numerator and denominator
Using L'Hôpital's rule, we differentiate the numerator and the denominator with respect to \(x\):

- **Numerator**:
\[ \frac{d}{dx}[\sin(ax) - \ln(e^x \cos x)] = a \cos(ax) - \left(\frac{1}{e^x \cos x} \cdot (e^x \cos x - e^x \sin x)\right) \]
Simplifying this gives:
\[ a \cos(ax) - \frac{e^x \cos x - e^x \sin x}{e^x \cos x} = a \cos(ax) - 1 + \tan x. \]

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

### Step 3: Rewrite the limit
Now we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{a \cos(ax) - 1 + \tan x}{\sin(bx) + bx \cos(bx)}. \]

### Step 4: Evaluate the limit at \(x = 0\)
Substituting \(x = 0\):
- The numerator becomes:
\[ a \cdot 1 - 1 + 0 = a - 1. \]
- The denominator becomes:
\[ 0 + 0 = 0. \]
This again gives us a \(0/0\) form, so we apply L'Hôpital's rule again.

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

### Step 6: Evaluate the new limit
Substituting \(x = 0\) again:
- The numerator becomes:
\[ -a^2 \cdot 0 + 1 = 1. \]
- The denominator becomes:
\[ b + 1. \]
Thus, we have:
\[ \lim_{x \to 0} \frac{1}{b + 1} = \frac{1}{2}. \]

### Step 7: Solve for \(b\)
Setting the limit equal to \(\frac{1}{2}\):
\[ \frac{1}{b + 1} = \frac{1}{2} \implies 2 = b + 1 \implies b = 1. \]

### Step 8: Find \(a\)
From earlier steps, we found that for the limit to exist, we need \(a - 1 = 0\), which gives:
\[ a = 1. \]

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

Thus, the value of \(a + b\) is \(\boxed{2}\).