Question:

Step 1: Find the left-hand limit as $x \to 0^-$: $$\lim_{x \to 0^-} \frac{\sin[a(x+1)] + \sin x}{2x}$$ As $x \to 0^-$, $a(x+1) \to a$, so $\sin[a(x+1)] \to \sin a$ (treating $[\cdot]$ as floor function, but here it seems $[a(x+1)]$ means the expression $a(x+1)$ without floor, or with floor). Assuming $[k]$ is the floor function: For $x$ near $0$, $a(x+1)$ is near $a$. If $a$ is not an integer, $[a(x+1)] = [a]$ for small $x$. So: $$\lim_{x \to 0^-} \frac{\sin([a]) + \sin x}{2x}$$ For this limit to be finite, $\sin([a]) = 0$, so $[a]$ must be a multiple of $\pi$... but $[a]$ is an integer, so $[a] = 0$, meaning $0 \leq a < 1$, or $[a] = n\pi$ which for integers means $[a]=0$. If $[a] = 0$: $\lim_{x\to 0^-} \dfrac{0 + \sin x}{2x} = \dfrac{1}{2}$. Step 2: Find the right-hand limit as $x \to 0^+$: $$\lim_{x \to 0^+} \frac{(x + bx^2)^{1/2} - x^{1/2}}{bx^{3/2}} = \lim_{x \to 0^+} \frac{x^{1/2}(1 + bx)^{1/2} - x^{1/2}}{bx^{3/2}}$$ $$= \lim_{x \to 0^+} \frac{(1+bx)^{1/2} - 1}{bx} = \lim_{x \to 0^+} \frac{1 + \frac{bx}{2} - 1}{bx} = \frac{1}{2}$$ Step 3: For continuity at $x=0$: LHL = $f(0)$ = RHL, so $c = \dfrac{1}{2}$. Step 4: Check option (3): $c = 1/2$ ✓ Step 5: Check option (1): $a = 5/3$. We need $[a] = 0$, i.e., $0 \leq a < 1$. Since $5/3 > 1$, $[5/3] = 1 \neq 0$. So $a = 5/3$ does NOT satisfy the continuity condition. ✗ Step 6: Check option (2): $b = 2$. The RHL is $1/2$ for any $b \neq 0$, so $b = 2$ is valid. ✓ Step 7: Check option (4): $f(1) = 1/3$. With $b = 2$: $$f(1) = \frac{(1 + 2)^{1/2} - 1}{2 \cdot 1^{3/2}} = \frac{\sqrt{3} - 1}{2} \approx \frac{0.732}{2} \approx 0.366 \neq \frac{1}{3}$$ For $f(1) = 1/3$: $\dfrac{(1+b)^{1/2} - 1}{b} = \dfrac{1}{3}$, so $3\sqrt{1+b} = b + 3$, squaring: $9(1+b) = (b+3)^2 = b^2 + 6b + 9$, giving $b^2 - 3b = 0$, so $b = 3$. With $b=3$: $f(1) = \dfrac{\sqrt{4}-1}{3} = \dfrac{1}{3}$ ✓ Step 8: The valid options are (2), (3), and (4).