Question:

Step 1: Expand $\sin 3x$ using the Taylor series. We use the Taylor expansion of sine: $\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - ...$ For $\sin 3x$: $$\sin 3x = 3x - \frac{(3x)^3}{6} + \frac{(3x)^5}{120} - ... = 3x - \frac{27x^3}{6} + ... = 3x - \frac{9x^3}{2} + ...$$ Step 2: Express $\frac{\sin 3x}{x^3}$ in terms of powers of $x$. Dividing the expansion by $x^3$: $$\frac{\sin 3x}{x^3} = \frac{3x - \frac{9x^3}{2} + ...}{x^3} = \frac{3}{x^2} - \frac{9}{2} + O(x^2)$$ Step 3: Substitute into the original limit expression. The given limit becomes: $$\lim_{x \to 0}\left(\frac{3}{x^2} - \frac{9}{2} + \frac{a}{x^2} + b\right) = \lim_{x \to 0}\left(\frac{3+a}{x^2} + \left(b - \frac{9}{2}\right) + ...\right)$$ Step 4: Apply the condition that the limit equals zero. For the limit to exist and equal $0$ (a finite value), the coefficient of $\frac{1}{x^2}$ must vanish, otherwise the expression would diverge as $x \to 0$. Therefore: $$3 + a = 0 \implies a = -3$$ Step 5: Find the value of $b$ using the limit condition. With $a = -3$, the expression becomes: $$\lim_{x \to 0}\left(b - \frac{9}{2} + ...\right) = 0$$ For this limit to equal zero: $$b - \frac{9}{2} = 0 \implies b = \frac{9}{2}$$ Step 6: Calculate $a + b$ and identify the answer. $$a + b = -3 + \frac{9}{2} = \frac{-6 + 9}{2} = \frac{3}{2}$$ The value of $(a+b) = \dfrac{3}{2}$, which corresponds to **Option 2**.