Key Concepts and Formulas

• Functional Equations: Solving for a function $f(x)$ given a relationship involving $f(x)$ and $f(1/x)$. This often involves substitution and solving a system of equations.
• Limits: Evaluating limits of functions, especially as $x$ approaches 0. This may involve algebraic manipulation to resolve indeterminate forms.
• Algebraic Manipulation: Techniques for simplifying expressions and solving equations.

Step-by-Step Solution

Step 1: Set up the given functional equation. We are given the functional equation: $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2} \quad \cdots (1)$ This equation relates the function $f(x)$ to its reciprocal counterpart $f(1/x)$.

Step 2: Substitute $x$ with $1/x$ in the functional equation. To solve for $f(x)$, we need another equation involving $f(x)$ and $f(1/x)$. We achieve this by replacing every instance of $x$ with $1/x$ in equation (1): $f\left(\frac{1}{x}\right)-6 f\left(\frac{1}{1/x}\right)=\frac{35}{3 (1/x)}-\frac{5}{2}$ Simplifying this gives: $f\left(\frac{1}{x}\right)-6 f(x)=\frac{35 x}{3}-\frac{5}{2} \quad \cdots (2)$

Step 3: Solve the system of linear equations for $f(x)$. Now we have a system of two linear equations with $f(x)$ and $f(1/x)$ as variables:

1. $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}$
2. $-6 f(x)+f\left(\frac{1}{x}\right)=\frac{35 x}{3}-\frac{5}{2}$

To eliminate $f(1/x)$, we can multiply equation (2) by 6 and add it to equation (1). Multiply equation (2) by 6: $6 \left(-6 f(x)+f\left(\frac{1}{x}\right)\right) = 6 \left(\frac{35 x}{3}-\frac{5}{2}\right)$ $-36 f(x)+6 f\left(\frac{1}{x}\right) = 70 x - 15 \quad \cdots (3)$

Now, add equation (1) and equation (3): $(f(x)-6 f\left(\frac{1}{x}\right)) + (-36 f(x)+6 f\left(\frac{1}{x}\right)) = \left(\frac{35}{3 x}-\frac{5}{2}\right) + (70 x - 15)$ $-35 f(x) = \frac{35}{3 x} + 70 x - \frac{5}{2} - 15$ $-35 f(x) = \frac{35}{3 x} + 70 x - \frac{35}{2}$

Divide by -35 to solve for $f(x)$: $f(x) = \frac{1}{-35} \left(\frac{35}{3 x} + 70 x - \frac{35}{2}\right)$ $f(x) = -\frac{1}{3 x} - 2 x + \frac{1}{2}$

Step 4: Evaluate the limit expression. We are given that $\lim\limits _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta$. Substitute the expression for $f(x)$ we found: $\lim\limits _{x \rightarrow 0}\left(\frac{1}{\alpha x} - 2 x - \frac{1}{3 x} + \frac{1}{2}\right) = \beta$ Rearrange the terms to group the $1/x$ terms: $\lim\limits _{x \rightarrow 0}\left(\left(\frac{1}{\alpha} - \frac{1}{3}\right) \frac{1}{x} - 2 x + \frac{1}{2}\right) = \beta$

Step 5: Determine the value of $\alpha$ for the limit to exist. For the limit to converge to a finite value $\beta$, the term with $1/x$ must vanish. This means the coefficient of $1/x$ must be zero. $\frac{1}{\alpha} - \frac{1}{3} = 0$ $\frac{1}{\alpha} = \frac{1}{3}$ $\alpha = 3$

Step 6: Determine the value of $\beta$. Now that we know $\alpha=3$, substitute this back into the limit expression and evaluate it: $\lim\limits _{x \rightarrow 0}\left(\left(\frac{1}{3} - \frac{1}{3}\right) \frac{1}{x} - 2 x + \frac{1}{2}\right) = \beta$ $\lim\limits _{x \rightarrow 0}\left(0 \cdot \frac{1}{x} - 2 x + \frac{1}{2}\right) = \beta$ $\lim\limits _{x \rightarrow 0}\left(- 2 x + \frac{1}{2}\right) = \beta$ As $x \rightarrow 0$, $-2x \rightarrow 0$. Therefore: $\beta = 0 + \frac{1}{2} = \frac{1}{2}$

Step 7: Calculate $\alpha + 2\beta$. We have found $\alpha = 3$ and $\beta = \frac{1}{2}$. Now, we compute $\alpha + 2\beta$: $\alpha + 2\beta = 3 + 2 \left(\frac{1}{2}\right)$ $\alpha + 2\beta = 3 + 1$ $\alpha + 2\beta = 4$

Common Mistakes & Tips

• Algebraic Errors: Be very careful when manipulating the equations, especially when multiplying or adding them. A small error can lead to an incorrect function $f(x)$ and thus an incorrect final answer.
• Limit of $1/x$ as $x \to 0$: Remember that $\lim\limits_{x \to 0} \frac{1}{x}$ does not exist (it tends to $\pm \infty$). For a limit involving $1/x$ to exist, the coefficient of $1/x$ must be zero.
• Substitution: Ensure that when substituting $x$ with $1/x$, all terms in the equation are correctly transformed. For example, $1/(1/x) = x$ and $1/(3(1/x)) = x/3$.

Summary

The problem involves solving a functional equation to find an explicit expression for $f(x)$. By substituting $x$ with $1/x$ and forming a system of linear equations, we successfully derived $f(x) = -2x - \frac{1}{3x} + \frac{1}{2}$. Subsequently, we used the given limit condition $\lim\limits _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta$. For this limit to exist, the term causing the divergence (the $1/x$ term) must have a zero coefficient, which allowed us to determine $\alpha = 3$. With $\alpha$ found, we evaluated the limit to find $\beta = 1/2$. Finally, we calculated $\alpha + 2\beta = 3 + 2(1/2) = 4$.

The final answer is \boxed{4}.