Limit Sinx Cosx C25F31

1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{1 + \sin x - \cos x}{1 - \sin x - \cos x}$$. 2. **Recall important formulas and approximations:** For small $x$, use the Taylor series approximations: $$\sin x \approx x - \frac{x^3}{6}$$ $$\cos x \approx 1 - \frac{x^2}{2}$$ 3. **Substitute approximations into numerator and denominator:** Numerator: $$1 + \sin x - \cos x \approx 1 + \left(x - \frac{x^3}{6}\right) - \left(1 - \frac{x^2}{2}\right) = 1 + x - \frac{x^3}{6} - 1 + \frac{x^2}{2} = x + \frac{x^2}{2} - \frac{x^3}{6}$$ Denominator: $$1 - \sin x - \cos x \approx 1 - \left(x - \frac{x^3}{6}\right) - \left(1 - \frac{x^2}{2}\right) = 1 - x + \frac{x^3}{6} - 1 + \frac{x^2}{2} = -x + \frac{x^2}{2} + \frac{x^3}{6}$$ 4. **Rewrite the limit using these approximations:** $$\lim_{x \to 0} \frac{x + \frac{x^2}{2} - \frac{x^3}{6}}{-x + \frac{x^2}{2} + \frac{x^3}{6}}$$ 5. **Factor out $x$ from numerator and denominator:** $$= \lim_{x \to 0} \frac{x\left(1 + \frac{x}{2} - \frac{x^2}{6}\right)}{x\left(-1 + \frac{x}{2} + \frac{x^2}{6}\right)}$$ 6. **Cancel the common factor $x$:** $$= \lim_{x \to 0} \frac{\cancel{x}\left(1 + \frac{x}{2} - \frac{x^2}{6}\right)}{\cancel{x}\left(-1 + \frac{x}{2} + \frac{x^2}{6}\right)}$$ 7. **Evaluate the limit by substituting $x=0$ in the simplified expression:** $$= \frac{1 + 0 - 0}{-1 + 0 + 0} = \frac{1}{-1} = -1$$ **Final answer:** $$\boxed{-1}$$