Question:

Step 1: Compute $\lim_{x \to 0^+} f(x)$. As $x \to 0^+$, $\{x\} \to 0^+$. Let $t = \{x\} \to 0^+$: $$\lim_{t \to 0^+} \frac{\cos^{-1}(1-t)\sin^{-1}(1-t)}{\sqrt{2t}(1-t)}$$ As $t \to 0^+$: $\sin^{-1}(1-t) \to \sin^{-1}(1) = \dfrac{\pi}{2}$ and $1-t \to 1$. For $\cos^{-1}(1-t)$ as $t \to 0^+$: use $\cos^{-1}(1-t) \approx \sqrt{2t}$. $$\lim_{t \to 0^+} \frac{\sqrt{2t} \cdot \frac{\pi}{2}}{\sqrt{2t} \cdot 1} = \frac{\pi}{2}$$ Step 2: Compute $\lim_{x \to 0^-} f(x)$. As $x \to 0^-$, $\{x\} \to 1^-$. Let $\{x\} = 1 - s$ where $s \to 0^+$: $$\lim_{s \to 0^+} \frac{\cos^{-1}(1-(1-s))\sin^{-1}(1-(1-s))}{\sqrt{2(1-s)} \cdot (1-(1-s))} = \lim_{s \to 0^+} \frac{\cos^{-1}(s)\sin^{-1}(s)}{\sqrt{2(1-s)} \cdot s}$$ As $s \to 0^+$: $\cos^{-1}(s) \to \dfrac{\pi}{2}$, $\sin^{-1}(s) \approx s$, $\sqrt{2(1-s)} \to \sqrt{2}$. $$= \lim_{s \to 0^+} \frac{\frac{\pi}{2} \cdot s}{\sqrt{2} \cdot s} = \frac{\pi}{2\sqrt{2}}$$ Step 3: Verify option (1): $\lim_{x \to 0^+} f(x) = \dfrac{\pi}{2}$ and $\sqrt{2} \lim_{x \to 0^-} f(x) = \sqrt{2} \cdot \dfrac{\pi}{2\sqrt{2}} = \dfrac{\pi}{2}$. So option (1) is correct. ✓ Step 4: Verify option (3): $\lim_{x \to 0^-} f(x) = \dfrac{\pi}{2\sqrt{2}}$. ✓