1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} \frac{1 - \cos x + \sin 3x}{1 + \tan 2x - \cos x}$$. 2. **Recall important formulas and approximations near zero:** - $\cos x \approx 1 - \frac{x^2}{2}$ - $\sin x \approx x$ - $\tan x \approx x$ 3. **Apply approximations to numerator:** $$1 - \cos x + \sin 3x \approx 1 - \left(1 - \frac{x^2}{2}\right) + 3x = \frac{x^2}{2} + 3x$$ 4. **Apply approximations to denominator:** $$1 + \tan 2x - \cos x \approx 1 + 2x - \left(1 - \frac{x^2}{2}\right) = 2x + \frac{x^2}{2}$$ 5. **Rewrite the limit using these approximations:** $$\lim_{x \to 0} \frac{\frac{x^2}{2} + 3x}{2x + \frac{x^2}{2}}$$ 6. **Factor out $x$ from numerator and denominator:** $$\lim_{x \to 0} \frac{x\left(\frac{x}{2} + 3\right)}{x\left(2 + \frac{x}{2}\right)} = \lim_{x \to 0} \frac{\frac{x}{2} + 3}{2 + \frac{x}{2}}$$ 7. **Evaluate the limit by substituting $x=0$:** $$\frac{0 + 3}{2 + 0} = \frac{3}{2}$$ **Final answer:** $\boxed{\frac{3}{2}}$ which corresponds to option (b).