Question:

Step 1: Verify that the limit is indeterminate. As $x \to 0^+$, we need to check the behavior of both numerator and denominator. **Numerator:** $\displaystyle\int_0^{\arctan x} \sin(t^2)\, dt \to \int_0^0 \sin(t^2)\, dt = 0$ **Denominator:** $x\cos x - x = x(\cos x - 1) \to 0 \cdot (1-1) = 0$ Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's rule or use Taylor series expansion. Step 2: Expand the numerator using Taylor series. For small values of $t$, we have: $$\sin(t^2) \approx t^2$$ Since $\arctan x \approx x$ for small $x$, we can approximate: $$\int_0^{\arctan x} \sin(t^2)\, dt \approx \int_0^x t^2\, dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3}$$ Step 3: Expand the denominator using Taylor series. For the denominator, we use the Taylor expansion $\cos x \approx 1 - \frac{x^2}{2}$ for small $x$: $$x\cos x - x = x\left(\cos x - 1\right) \approx x\left(1 - \frac{x^2}{2} - 1\right) = x \cdot \left(-\frac{x^2}{2}\right) = -\frac{x^3}{2}$$ Step 4: Compute the limit. Substituting the approximations from Steps 2 and 3: $$\lim_{x \to 0^+} \frac{\displaystyle\int_0^{\arctan x} \sin(t^2)\, dt}{x\cos x - x} = \lim_{x \to 0^+} \frac{\frac{x^3}{3}}{-\frac{x^3}{2}}$$ Simplifying: $$= \frac{\frac{1}{3}}{-\frac{1}{2}} = \frac{1}{3} \cdot \frac{-2}{1} = -\frac{2}{3}$$ **Final Answer:** The value of the limit is $\boxed{-\dfrac{2}{3}}$, which corresponds to **Option 4**.