1. The problem asks to show that when $x$ is very small such that terms involving $x^4$ and higher powers can be neglected, $$\frac{e^x}{1+x} = 1 + \frac{x^2}{2} - \frac{x^3}{3}.$$ 2. Start by writing the Taylor expansion of $e^x$ around $x=0$: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$ 3. Since $x$ is very small, neglect $x^4$ and higher powers, so approximately: $$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}.$$ 4. The denominator is $1 + x$, so write the expression: $$\frac{e^x}{1+x} \approx \frac{1 + x + \frac{x^2}{2} + \frac{x^3}{6}}{1 + x}.$$ 5. Divide numerator by denominator using polynomial division or consider the series expansion for the quotient: Assume $$\frac{e^x}{1+x} = A + Bx + Cx^2 + Dx^3.$$ 6. Multiply both sides by $(1 + x)$ to match numerator: $$(A + Bx + Cx^2 + Dx^3)(1 + x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}.$$ 7. Expand left side: $$A + Ax + Bx + Bx^2 + Cx^2 + Cx^3 + Dx^3 + Dx^4.$$ Neglect $x^4$ and higher: $$A + (A + B)x + (B + C)x^2 + (C + D)x^3.$$ 8. Equate coefficients of powers of $x$ from left and right sides: - Constant: $$A = 1$$ - Coefficient of $x$: $$A + B = 1$$ - Coefficient of $x^2$: $$B + C = \frac{1}{2}$$ - Coefficient of $x^3$: $$C + D = \frac{1}{6}.$$ 9. From $A=1$, then $1 + B = 1 \Rightarrow B=0.$ 10. From $B + C = \frac{1}{2}$ and $B=0$, get $C=\frac{1}{2}.$ 11. From $C + D = \frac{1}{6}$, get $\frac{1}{2} + D = \frac{1}{6} \Rightarrow D = \frac{1}{6} - \frac{1}{2} = -\frac{1}{3}.$ 12. Substitute $A,B,C,D$ back: $$\frac{e^x}{1+x} \approx 1 + 0 \cdot x + \frac{x^2}{2} - \frac{x^3}{3} = 1 + \frac{x^2}{2} - \frac{x^3}{3}.$$ This shows the desired result.