Let f(x) = begin{cases} alpha + frac{sin [x]} | vedclass

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(B) Given that $f(x)$ is continuous at $x = 0$, the condition $\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$ must hold. First, calculate the L

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(B) Given that $f(x)$ is continuous at $x = 0$, the condition $\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$ must hold. First, calculate the Left-Hand Limit $(LHL)$: $	ext{LHL} = \lim_{x 	o 0^-} \left( \beta + \left[ \frac{\sin x - x}{x^3} ight] ight)$. Using the Taylor series expansion, $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$, so $\frac{\sin x - x}{x^3} = \frac{-x^3/6 + x^5/120}{x^3} = -\frac{1}{6} + \frac{x^2}{120}$. As $x 	o 0^-$, $\frac{\sin x - x}{x^3} 	o -\frac{1}{6}$. Thus, $	ext{LHL} = \beta + [-1/6] = \beta - 1$. Next, calculate the Right-Hand Limit $(RHL)$: $	ext{RHL} = \lim_{x 	o 0^+} \left( \alpha + \frac{\sin [x]}{x} ight)$. For $0 Thus, $	ext{RHL} = \lim_{x 	o 0^+} (\alpha + 0) = \alpha$. Given $f(0) = 2$, we have $\beta - 1 = 2 = \alpha$. So, $\beta = 3$ and $\alpha = 2$. Therefore, $\beta - \alpha = 3 - 2 = 1$.

Let $f(x) = \begin{cases} \alpha + \frac{\sin [x]}{x}, & \text{if } x > 0 \\ 2, & \text{if } x = 0 \\ \beta + \left[ \frac{\sin x - x}{x^3} \right], & \text{if } x < 0 \end{cases}$ where $[x]$ denotes the greatest integer function. If $f$ is continuous at $x = 0$, then $\beta - \alpha$ is equal to

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