Evaluation of Limit Involving Trigonometric Functions and Constants

To evaluate the given limit and find the value of $6(\alpha + \beta)$, we use the Taylor series expansion of the sine function around $x = 0$.

The Taylor series expansion for $\sin(t)$ is given by: $\sin(t) = t - \frac{t^3}{6} + O(t^5)$

Applying this expansion to $\sin(\beta x)$ and $\sin x$, we get: $\sin(\beta x) = \beta x - \frac{\beta^3 x^3}{6} + O(x^5)$ $\sin x = x - \frac{x^3}{6} + O(x^5)$

Substitute these expansions into the given limit expression: $\lim_{x \to 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = \lim_{x \to 0} \frac{x^2 \left( \beta x - \frac{\beta^3 x^3}{6} + O(x^5) \right)}{\alpha x - \left( x - \frac{x^3}{6} + O(x^5) \right)}$

$\lim_{x \to 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + O(x^7)}{(\alpha - 1)x + \frac{x^3}{6} + O(x^5)} = 1$

For this limit to exist and be a non-zero finite value ($1$), the lowest power of $x$ in the denominator must equal the lowest power of $x$ in the numerator, which is $x^3$. Thus, the coefficient of $x$ in the denominator must be zero: $\alpha - 1 = 0 \implies \alpha = 1$

With $\alpha = 1$, the denominator simplifies to $\frac{x^3}{6} + O(x^5)$, and the limit becomes: $\lim_{x \to 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + O(x^7)}{\frac{x^3}{6} + O(x^5)} = 1$

Divide both numerator and denominator by $x^3$: $\lim_{x \to 0} \frac{\beta - \frac{\beta^3 x^2}{6} + O(x^4)}{\frac{1}{6} + O(x^2)} = 1$

Taking the limit as $x \to 0$: $\frac{\beta}{\frac{1}{6}} = 1 \implies 6\beta = 1 \implies \beta = \frac{1}{6}$

Now, we calculate $6(\alpha + \beta)$: $6(\alpha + \beta) = 6 \left( 1 + \frac{1}{6} \right) = 6 + 1 = 7$