Let $f: \mathbb{R} - \{0\} \rightarrow \mathbb{R}$ be a function such that $f(x) - 6 f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If the $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$; $\alpha, \beta \in \mathbb{R}$, then $\alpha + 2\beta$ is equal to

Step 1: Formulate a system of equations for f(x)

We are given an equation involving $f(x)$ and $f(1/x)$. To solve for $f(x)$, we can create a system of two linear equations. The first equation is the given one. The second equation is obtained by replacing $x$ with $1/x$ in the original equation. This substitution transforms $f(x)$ into $f(1/x)$ and $f(1/x)$ into $f(x)$.

Step 2: Solve for f(x)

To solve for $f(x)$, we multiply equation (2) by 6 and add it to equation (1). This eliminates $f(1/x)$, allowing us to isolate $f(x)$. After combining terms and simplifying, we find the explicit expression for $f(x)$.

Step 3: Evaluate the limit

Now we substitute the derived expression for $f(x)$ into the given limit expression. The limit is given as $\beta$. We need to analyze this limit as $x$ approaches 0.

Step 4: Determine $\alpha$ and $\beta$

For the limit to exist and be finite (equal to $\beta$), the term with $1/x$ must vanish. This means the coefficient of $1/x$ must be zero. From this, we can find $\alpha$. Once $\alpha$ is found, the limit simplifies to the remaining terms as $x \rightarrow 0$, which gives us $\beta$.

Step 5: Calculate $\alpha + 2\beta$

From the previous step, setting the coefficient of $1/x$ to zero gives $1/\alpha + 1/3 = 0$, which implies $\alpha = -3$. Then, substituting $\alpha = -3$ into the limit expression, we get $\lim_{x \rightarrow 0} (6x - 1/2) = -1/2$, so $\beta = -1/2$. Finally, we calculate $\alpha + 2\beta$.

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