Question

If \[ \lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log(1 - x)}{3 \tan^2 x} = \frac{1}{3}, \] then \( 2\alpha - \beta \) is equal to:

• A. 2
• B. 7
• C. 5
• D. 1

Answer 1

The Correct Option is C

Given:
\(\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1 - x)}{3 \tan^2 x} = \frac{1}{3}.\)

Expanding the trigonometric and logarithmic functions around \(x = 0\) using Taylor series:  
\(\sin x \approx x, \quad \cos x \approx 1 - \frac{x^2}{2}, \quad \log_e(1 - x) \approx -x - \frac{x^2}{2}.\)

Substituting these approximations:  
\(\lim_{x \to 0} \frac{3 + \alpha x + \beta \left(1 - \frac{x^2}{2}\right) - x - \frac{x^2}{2}}{3 \left(\frac{x^3}{3} + \ldots\right)^2}.\)

Simplifying the numerator:  
\(3 + \beta + (\alpha - 1)x + \left(-\frac{\beta}{2} - \frac{1}{2}\right)x^2.\)

For the limit to be finite and equal to \(\frac{1}{3}\), the terms involving \(x\) and higher powers of \(x\) must vanish. This gives:  
\(\alpha - 1 = 0 \implies \alpha = 1,\)
and:  
\(3 + \beta = 0 \implies \beta = -3.\)

Substituting these values:  
\(2\alpha - \beta = 2 \times 1 - (-3) = 2 + 3 = 5.\)

Thus, the correct answer is: 5