Let `alpha,betainR` be such that `lim_(xto0) (x^(2)sin(betax))/(alphax-sinx)=1`. Then `6(alpha+beta)` equals___________.

Explanation

To solve the limit problem, we start with the given expression:

\[ \lim_{x \to 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1 \]

### Step 1: Rewrite the limit expression
We can rewrite the limit by dividing both the numerator and the denominator by \(\beta x\):

\[ \lim_{x \to 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = \lim_{x \to 0} \frac{x^2}{\beta x} \cdot \frac{\sin(\beta x)}{\beta x} \cdot \frac{\beta x}{\alpha x - \sin x} \]

### Step 2: Use the limit property
Using the property that \(\lim_{x \to 0} \frac{\sin(\beta x)}{\beta x} = 1\), we simplify the limit:

\[ = \lim_{x \to 0} \frac{x}{\beta} \cdot 1 \cdot \frac{\beta x}{\alpha x - \sin x} \]

### Step 3: Substitute the Taylor series for \(\sin x\)
Using the Taylor series expansion for \(\sin x\):

\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

We substitute this into the denominator:

\[ \alpha x - \sin x = \alpha x - \left(x - \frac{x^3}{6} + O(x^5)\right) = (\alpha - 1)x + \frac{x^3}{6} - O(x^5) \]

### Step 4: Substitute back into the limit
Now we substitute this back into our limit:

\[ \lim_{x \to 0} \frac{x^2 \cdot \frac{\sin(\beta x)}{\beta x}}{\alpha x - \sin x} = \lim_{x \to 0} \frac{x^2}{\beta} \cdot \frac{\beta x}{(\alpha - 1)x + \frac{x^3}{6}} = \lim_{x \to 0} \frac{x^2}{\beta} \cdot \frac{\beta x}{(\alpha - 1)x + \frac{x^3}{6}} \]

### Step 5: Simplify the limit
As \(x\) approaches 0, we can ignore the higher-order terms:

\[ = \lim_{x \to 0} \frac{x^2}{\beta} \cdot \frac{\beta x}{(\alpha - 1)x} = \lim_{x \to 0} \frac{x^3}{\beta(\alpha - 1)} \]

### Step 6: Set the limit equal to 1
For the limit to equal 1, we need:

\[ \frac{1}{\beta(\alpha - 1)} = 1 \implies \beta(\alpha - 1) = 1 \]

### Step 7: Solve for \(\alpha\)
From the equation \(\beta(\alpha - 1) = 1\):

\[ \alpha - 1 = \frac{1}{\beta} \implies \alpha = 1 + \frac{1}{\beta} \]

### Step 8: Substitute \(\alpha\) back into the equation
Now we can substitute \(\alpha\) back into the equation:

\[ \beta \left( 1 + \frac{1}{\beta} - 1 \right) = 1 \implies \beta \cdot \frac{1}{\beta} = 1 \implies \beta = \frac{1}{6} \]

### Step 9: Find \(\alpha\)
Substituting \(\beta = \frac{1}{6}\) back into the equation for \(\alpha\):

\[ \alpha = 1 + 6 = 7 \]

### Step 10: Calculate \(6(\alpha + \beta)\)
Now we can calculate \(6(\alpha + \beta)\):

\[ 6(\alpha + \beta) = 6\left(7 + \frac{1}{6}\right) = 6\left(\frac{42 + 1}{6}\right) = 6 \cdot 7 = 42 \]

Thus, the final answer is:

\[ \boxed{42} \]