Key Concepts and Formulas

• Standard Limits: We will use the following standard limits:
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{\tan x}{x} = 1$
  • $\lim_{x \to 0} \frac{\log_e(1+x)}{x} = 1$
• Taylor Series Expansions (around x=0): For small $x$, we can use the following Taylor series expansions:
  • $\sin x = x - \frac{x^3}{3!} + O(x^5)$
  • $\cos x = 1 - \frac{x^2}{2!} + O(x^4)$
  • $\log_e(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)$
  • $\tan x = x + \frac{x^3}{3} + O(x^5)$
• Indeterminate Forms: If the limit results in an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$), we can use Taylor series expansions or L'Hôpital's Rule to evaluate it.

Step-by-Step Solution

Step 1: Analyze the given limit and identify the indeterminate form. The given limit is: $\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e (1 - x)}{3 \tan^2 x} = \frac{1}{3}$ As $x \to 0$, the numerator approaches $3 + \alpha(0) + \beta(1) + \log_e(1) = 3 + \beta$. The denominator approaches $3 \tan^2(0) = 3(0)^2 = 0$. For the limit to be a finite non-zero value ($\frac{1}{3}$), the numerator must also approach 0. Therefore, we must have $3 + \beta = 0$, which implies $\beta = -3$.

Step 2: Substitute the value of $\beta$ and use Taylor series expansions for the numerator and denominator. Now that we know $\beta = -3$, the numerator becomes: $3 + \alpha \sin x - 3 \cos x + \log_e (1 - x)$ Using the Taylor series expansions around $x=0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$ $\cos x = 1 - \frac{x^2}{2} + O(x^4)$ $\log_e (1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)$ The numerator is: $3 + \alpha \left(x - \frac{x^3}{6} + O(x^5)\right) - 3 \left(1 - \frac{x^2}{2} + O(x^4)\right) + \left(-x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)\right)$ $= 3 + \alpha x - \frac{\alpha x^3}{6} - 3 + \frac{3x^2}{2} - x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)$ Combine terms: $= (\alpha - 1)x + \left(\frac{3}{2} - \frac{1}{2}\right)x^2 + \left(-\frac{\alpha}{6} - \frac{1}{3}\right)x^3 + O(x^4)$ $= (\alpha - 1)x + x^2 + \left(-\frac{\alpha}{6} - \frac{2}{6}\right)x^3 + O(x^4)$ $= (\alpha - 1)x + x^2 - \frac{\alpha + 2}{6}x^3 + O(x^4)$

For the denominator: $\tan x = x + \frac{x^3}{3} + O(x^5)$ $\tan^2 x = \left(x + \frac{x^3}{3} + O(x^5)\right)^2 = x^2 + 2x\left(\frac{x^3}{3}\right) + O(x^6) = x^2 + \frac{2x^4}{3} + O(x^6)$ So, $3 \tan^2 x = 3 \left(x^2 + O(x^4)\right) = 3x^2 + O(x^4)$.

Step 3: Rewrite the limit using the expanded forms and equate it to the given value. The limit becomes: $\lim_{x \to 0} \frac{(\alpha - 1)x + x^2 - \frac{\alpha + 2}{6}x^3 + O(x^4)}{3x^2 + O(x^4)} = \frac{1}{3}$ For this limit to be finite, the lowest power of $x$ in the numerator must be at least equal to the lowest power of $x$ in the denominator, which is $x^2$. This means the coefficient of the $x$ term in the numerator must be zero. So, $\alpha - 1 = 0$, which implies $\alpha = 1$.

Step 4: Substitute the value of $\alpha$ and re-evaluate the limit to find the coefficient of $x^2$. With $\alpha = 1$, the numerator becomes: $(\alpha - 1)x + x^2 - \frac{\alpha + 2}{6}x^3 + O(x^4) = (1 - 1)x + x^2 - \frac{1 + 2}{6}x^3 + O(x^4)$ $= 0x + x^2 - \frac{3}{6}x^3 + O(x^4)$ $= x^2 - \frac{1}{2}x^3 + O(x^4)$

The limit expression is now: $\lim_{x \to 0} \frac{x^2 - \frac{1}{2}x^3 + O(x^4)}{3x^2 + O(x^4)}$ Divide both numerator and denominator by $x^2$: $\lim_{x \to 0} \frac{1 - \frac{1}{2}x + O(x^2)}{3 + O(x^2)}$ Now, substitute $x=0$: $\frac{1 - 0 + 0}{3 + 0} = \frac{1}{3}$ This matches the given limit value, confirming our values of $\alpha = 1$ and $\beta = -3$.

Step 5: Calculate the value of $2\alpha - \beta$. We found $\alpha = 1$ and $\beta = -3$. Therefore, $2\alpha - \beta = 2(1) - (-3) = 2 + 3 = 5$.

Common Mistakes & Tips

• Incorrectly assuming the numerator is zero: If the numerator doesn't approach zero when the denominator does, the limit would be $\infty$ or $-\infty$, not a finite value. Always check the numerator's behavior first.
• Errors in Taylor series expansion: Ensure accuracy when expanding functions like $\sin x$, $\cos x$, $\log_e(1-x)$, and $\tan x$. Pay attention to the signs and coefficients.
• Algebraic mistakes: Be careful with combining terms after expansion, especially when dealing with coefficients and signs.

Summary The problem requires evaluating a limit that results in an indeterminate form. We first analyze the numerator and denominator as $x \to 0$. For the limit to be finite, the numerator must also be zero, which helps us determine the value of $\beta$. Then, using Taylor series expansions for the trigonometric and logarithmic functions, we simplify the numerator and denominator. By ensuring the lowest power of $x$ in the numerator is at least that of the denominator, we find the value of $\alpha$. Finally, we re-evaluate the limit with the determined values of $\alpha$ and $\beta$ to confirm consistency and then compute the required expression $2\alpha - \beta$.

Final Answer The final answer is $\boxed{5}$.