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(B) Given that $f(x)$ is continuous at $x=0$,therefore $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 2$. First,find the Left

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(B) Given that $f(x)$ is continuous at $x=0$,therefore $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 2$. First,find the Left Hand Limit $(LHL)$: $\lim_{x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(-h) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{\sin(-h) - (-h)}{(-h)^3} \right] \right) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{-\sin h + h}{-h^3} \right] \right) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{\sin h - h}{h^3} \right] \right)$. Using the Taylor series expansion $\sin h = h - \frac{h^3}{6} + \dots$,we get $\frac{\sin h - h}{h^3} = \frac{-h^3/6}{h^3} = -\frac{1}{6}$. Thus,$LHL$ $= \beta + [-1/6] = \beta - 1$. Since $LHL$ $= f(0)$,we have $\beta - 1 = 2 \Rightarrow \beta = 3$. Next,find the Right Hand Limit $(RHL)$: $\lim_{x \rightarrow 0^+} f(x) = \lim_{h \rightarrow 0} \left( \alpha + \frac{\sin [h]}{h} \right)$. For $0 Since $RHL$ $= f(0)$,we have $\alpha = 2$. Finally,$\beta - \alpha = 3 - 2 = 1$.

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