`lim_(xto0)(3-asinx-bcosx-log_e(1+x))/(3tan^2x)` is non zero finite find `2b-a`

Options

A. 2
B. 5
C. 7
D. 9

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{3 - a \sin x - b \cos x - \log(1+x)}{3 \tan^2 x} \]

we need to evaluate the limit and find the values of \(a\) and \(b\) such that the limit is non-zero and finite.

### Step-by-Step Solution:

1. **Expand the Functions**:
- We start by expanding \(\sin x\), \(\cos x\), and \(\log(1+x)\) using Taylor series around \(x=0\):
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
\[ \log(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4) \]

2. **Substitute the Expansions**:
- Substitute these expansions into the limit expression:
\[ 3 - a\left(x - \frac{x^3}{6}\right) - b\left(1 - \frac{x^2}{2}\right) - \left(x - \frac{x^2}{2} + \frac{x^3}{3}\right) \]
- This simplifies to:
\[ 3 - ax + \frac{ax^3}{6} - b + \frac{bx^2}{2} - x + \frac{x^2}{2} - \frac{x^3}{3} \]
- Combine like terms:
\[ (3 - b) + \left(-a - 1 + \frac{b}{2}\right)x + \left(\frac{a}{6} - \frac{1}{3}\right)x^3 + O(x^4) \]

3. **Expand the Denominator**:
- The denominator \(3 \tan^2 x\) can be expanded as:
\[ \tan x \approx x + \frac{x^3}{3} + O(x^5) \Rightarrow \tan^2 x \approx x^2 + \frac{2x^4}{3} + O(x^6) \]
- Thus,
\[ 3 \tan^2 x \approx 3x^2 + 2x^4 + O(x^6) \]

4. **Set Up the Limit**:
- Now we can write the limit:
\[ \lim_{x \to 0} \frac{(3 - b) + \left(-a - 1 + \frac{b}{2}\right)x + \left(\frac{a}{6} - \frac{1}{3}\right)x^3 + O(x^4)}{3x^2 + 2x^4 + O(x^6)} \]

5. **Evaluate the Limit**:
- For the limit to be finite and non-zero, the constant term in the numerator must be zero:
\[ 3 - b = 0 \Rightarrow b = 3 \]
- The coefficient of \(x\) must also be zero:
\[ -a - 1 + \frac{3}{2} = 0 \Rightarrow -a + \frac{1}{2} = 0 \Rightarrow a = \frac{1}{2} \]

6. **Find \(2b - a\)**:
- Now, substituting the values of \(a\) and \(b\):
\[ 2b - a = 2(3) - \frac{1}{2} = 6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2} \]

### Final Answer:
\[ 2b - a = \frac{11}{2} \]