1. **Problem 1:** Evaluate $$\lim_{x \to 0} \frac{x e^x - \ln(1+x)}{x^2}$$ 2. **Recall the expansions:** - Exponential function: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ - Natural logarithm: $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$ 3. **Substitute expansions into the numerator:** $$x e^x = x \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) = x + x^2 + \frac{x^3}{2} + \frac{x^4}{6}$$ 4. **Write numerator:** $$x e^x - \ln(1+x) = \left(x + x^2 + \frac{x^3}{2} + \frac{x^4}{6}\right) - \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\right)$$ 5. **Simplify numerator:** $$= x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4}$$ $$= \left(x - x\right) + \left(x^2 + \frac{x^2}{2}\right) + \left(\frac{x^3}{2} - \frac{x^3}{3}\right) + \left(\frac{x^4}{6} + \frac{x^4}{4}\right)$$ $$= 0 + \frac{3x^2}{2} + \frac{x^3}{6} + \frac{5x^4}{12}$$ 6. **Divide numerator by $x^2$:** $$\frac{x e^x - \ln(1+x)}{x^2} = \frac{\frac{3x^2}{2} + \frac{x^3}{6} + \frac{5x^4}{12}}{x^2} = \frac{3}{2} + \frac{x}{6} + \frac{5x^2}{12}$$ 7. **Take the limit as $x \to 0$:** $$\lim_{x \to 0} \left(\frac{3}{2} + \frac{x}{6} + \frac{5x^2}{12}\right) = \frac{3}{2}$$ --- 8. **Problem 2:** Evaluate $$\lim_{x \to 0} \frac{e^x - e^{-x}}{x}$$ 9. **Recall expansions:** - $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ - $$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$ 10. **Calculate numerator:** $$e^x - e^{-x} = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) - \left(1 - x + \frac{x^2}{2} - \frac{x^3}{6}\right) = x + x + \frac{x^3}{6} + \frac{x^3}{6} = 2x + \frac{x^3}{3}$$ 11. **Divide by $x$:** $$\frac{e^x - e^{-x}}{x} = \frac{2x + \frac{x^3}{3}}{x} = 2 + \frac{x^2}{3}$$ 12. **Take the limit as $x \to 0$:** $$\lim_{x \to 0} \left(2 + \frac{x^2}{3}\right) = 2$$ **Final answers:** - Problem 1 limit = $$\frac{3}{2}$$ - Problem 2 limit = $$2$$