The values of a,b and c, such that `lim_(x rarr 0) (ae^(x)- bcos x + c e^(-x))/(x sin x ) ` = 2 are

Options

A. a = 1,b = − 2, c = 1
B. a = 1,b = 2, c = − 1
C. a = 1,b = 2, c = 1
D. a = − 1,b = 2, c = 1

Answer

To solve the limit problem given by

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

we will use Taylor series expansions for \(e^x\), \(\cos x\), and \(\sin x\) around \(x = 0\).

### Step 1: Expand the functions using Taylor series

1. **Expansion of \(e^x\)**:
\[ e^x = 1 + x + \frac{x^2}{2} + O(x^3) \]

2. **Expansion of \(\cos x\)**:
\[ \cos x = 1 - \frac{x^2}{2} + O(x^4) \]

3. **Expansion of \(e^{-x}\)**:
\[ e^{-x} = 1 - x + \frac{x^2}{2} + O(x^3) \]

4. **Expansion of \(\sin x\)**:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

### Step 2: Substitute the expansions into the limit expression

Substituting the expansions into the limit expression gives:

\[ ae^x - b\cos x + c e^{-x} = a\left(1 + x + \frac{x^2}{2}\right) - b\left(1 - \frac{x^2}{2}\right) + c\left(1 - x + \frac{x^2}{2}\right) \]

Combining these, we have:

\[ = (a - b + c) + (a - c)x + \left(a + b + c\right)\frac{x^2}{2} + O(x^3) \]

### Step 3: Expand the denominator

The denominator \(x \sin x\) becomes:

\[ x \sin x = x\left(x - \frac{x^3}{6} + O(x^5)\right) = x^2 - \frac{x^4}{6} + O(x^6) \]

### Step 4: Formulate the limit

Now, substituting the numerator and denominator into the limit gives:

\[ \lim_{x \to 0} \frac{(a - b + c) + (a - c)x + \left(a + b + c\right)\frac{x^2}{2}}{x^2 - \frac{x^4}{6} + O(x^6)} \]

### Step 5: Analyze the limit

As \(x \to 0\), for the limit to exist and equal to 2, the constant term in the numerator must be zero:

1. **Setting the constant term to zero**:
\[ a - b + c = 0 \quad \text{(1)} \]

2. **Setting the coefficient of \(x\) to zero**:
\[ a - c = 0 \quad \text{(2)} \]

3. **Setting the coefficient of \(x^2\)**:
\[ a + b + c = 4 \quad \text{(3)} \quad \text{(since the limit equals 2)} \]

### Step 6: Solve the equations

From equation (2), we have:

\[ a = c \]

Substituting \(c\) with \(a\) in equation (1):

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

Now substituting \(b\) and \(c\) into equation (3):

\[ a + 2a + a = 4 \implies 4a = 4 \implies a = 1 \]

Thus, we find:

\[ c = a = 1, \quad b = 2a = 2 \]

### Final Values

The values of \(a\), \(b\), and \(c\) are:

\[ a = 1, \quad b = 2, \quad c = 1 \]