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

Answer: (C)

5

Answer 2

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