Question:

Step 1: Expand the numerator using Taylor series. We need to find the Taylor expansion of $\sqrt{1+x^2}$ around $x=0$: $$\sqrt{1+x^2} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + O(x^6)$$ Therefore, the numerator becomes: $$\frac{x^2}{2} + 1 - \sqrt{1+x^2} = \frac{x^2}{2} + 1 - \left(1 + \frac{x^2}{2} - \frac{x^4}{8} + O(x^6)\right)$$ Simplifying: $$= \frac{x^2}{2} + 1 - 1 - \frac{x^2}{2} + \frac{x^4}{8} + O(x^6) = \frac{x^4}{8} + O(x^6)$$ Step 2: Expand the first factor of the denominator using Taylor series. We need the Taylor expansions of $\cos x$ and $e^{x^2}$: $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$ $$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + O(x^6)$$ Therefore: $$\cos x - e^{x^2} = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - \left(1 + x^2 + \frac{x^4}{2}\right) + O(x^6)$$ $$= -\frac{x^2}{2} - x^2 + \frac{x^4}{24} - \frac{x^4}{2} + O(x^6)$$ $$= -\frac{3x^2}{2} - \frac{11x^4}{24} + O(x^6)$$ Step 3: Expand the second factor of the denominator using Taylor series. For small $x$, the Taylor expansion of $\sin(x^2)$ is: $$\sin(x^2) = x^2 - \frac{x^6}{6} + O(x^{10})$$ To leading order, $\sin(x^2) \approx x^2$. Step 4: Find the leading term of the denominator. The denominator is: $$(\cos x - e^{x^2})\sin(x^2) = \left(-\frac{3x^2}{2} + O(x^4)\right)(x^2 + O(x^6))$$ $$= -\frac{3x^2}{2} \cdot x^2 + O(x^6) = -\frac{3x^4}{2} + O(x^6)$$ Step 5: Calculate the limit by dividing the leading terms. $$\lim_{x \to 0} \frac{\frac{x^4}{8} + O(x^6)}{-\frac{3x^4}{2} + O(x^6)} = \frac{\frac{x^4}{8}}{-\frac{3x^4}{2}}$$ $$= \frac{1}{8} \cdot \frac{-2}{3} = -\frac{2}{24} = -\frac{1}{12}$$ **Final Answer:** The value of the limit is $\boxed{-\dfrac{1}{12}}$, which corresponds to **Option 2**.