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

To solve the given problem, we need to find \( 2\alpha - \beta \) so that the following limit holds:

\[ \lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log(1 - x)}{3 \tan^2 x} = \frac{1}{3}. \]

1. First, we simplify the expressions involved around \( x \to 0 \):
   • As \( x \to 0, \sin x \approx x \), \(\tan x \approx x\), and \(\cos x \approx 1\).
   • Using the Taylor series, \(\log(1 - x) \approx -x\) for small \( x \).
2. Substitute these approximations into the limit expression:
   • The numerator becomes:
     \(3 + \alpha x + \beta \cdot 1 - x = 3 + (\alpha - 1)x + \beta\).
   • The denominator simplifies using \(\tan^2 x \approx x^2\):
     \(3 \tan^2 x \approx 3x^2\).
   • Thus, the simplified expression for the limit is:
     \[ \lim_{x \to 0} \frac{3 + (\alpha - 1)x + \beta}{3x^2}. \]
3. For the limit to exist and equal \(\frac{1}{3}\), the constant term must vanish as the non-zero constant over \(x^2\) would diverge:
   • Therefore, both \(\alpha - 1\) and \(\beta\) should be zero, simplifying to:
     • \(3 + \beta = 0 \implies \beta = -3\).
     • \(\alpha - 1 = 0 \implies \alpha = 1\).
4. Calculate \(2\alpha - \beta\):
   \[ 2\alpha - \beta = 2(1) - (-3) = 2 + 3 = 5. \]
5. Thus, the value of \(2\alpha - \beta\) is 5.

Hence, the correct answer is 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