To solve the given limit:

$\lim_{x \to 0} \frac{\sin(x - 2x^2) - \cos(x) + e^{-x + x^2}}{4x^2 \ln(1-x)} = \frac{1}{3},$

we proceed step by step, expanding each term using Taylor series approximations around $x = 0$ and analyzing the dominant behavior of the numerator and denominator.

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Step 1: Taylor series for the numerator

1. For $\sin(x - 2x^2)$: $\sin(x - 2x^2) \approx \sin(x) - 2x^2 \cos(x).$ Substituting the Taylor series for $\sin(x)$ and $\cos(x)$ at $x = 0$: $\sin(x) \approx x - \frac{x^3}{6}, \quad \cos(x) \approx 1 - \frac{x^2}{2}.$ So: $\sin(x - 2x^2) \approx x - \frac{x^3}{6} - 2x^2(1 - \frac{x^2}{2}) \approx x - 2x^2 - \frac{x^3}{6} + x^4.$

2. For $-\cos(x)$: Using the series for $\cos(x)$: $-\cos(x) \approx -(1 - \frac{x^2}{2} + \frac{x^4}{24}) = -1 + \frac{x^2}{2} - \frac{x^4}{24}.$

3. For $e^{-x + x^2}$: Using the series for $e^z$, where $z = -x + x^2$: $e^{-x + x^2} \approx 1 - (x - x^2) + \frac{(x - x^2)^2}{2}.$ Expanding: $e^{-x + x^2} \approx 1 - x + x^2 + \frac{x^2 - 2x^3 + x^4}{2} = 1 - x + \frac{3x^2}{2} - x^3 + \frac{x^4}{2}.$

Combining the terms: The numerator is: $\sin(x - 2x^2) - \cos(x) + e^{-x + x^2} \approx \left(x - 2x^2 - \frac{x^3}{6} + x^4 \right) + \left(-1 + \frac{x^2}{2} - \frac{x^4}{24}\right) + \left(1 - x + \frac{3x^2}{2} - x^3 + \frac{x^4}{2}\right).$ Simplify: $\text{Numerator} \approx x - x + (-2x^2 + \frac{x^2}{2} + \frac{3x^2}{2}) + \left(-\frac{x^3}{6} - x^3 \right) + \left(x^4 + \frac{x^4}{2} - \frac{x^4}{24}\right).$ $\text{Numerator} \approx 2x^2 - \frac{7x^3}{6} + \frac{35x^4}{24}.$

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Step 2: Taylor series for the denominator

1. For $\ln(1-x)$: Using the series for $\ln(1-x)$: $\ln(1-x) \approx -x - \frac{x^2}{2} - \frac{x^3}{3}.$ Multiply by $4x^2$: $4x^2 \ln(1-x) \approx 4x^2(-x - \frac{x^2}{2} - \frac{x^3}{3}) = -4x^3 - 2x^4 - \frac{4x^5}{3}.$

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Step 3: Simplify the fraction

The fraction becomes: $\frac{\text{Numerator}}{\text{Denominator}} \approx \frac{2x^2 - \frac{7x^3}{6} + \frac{35x^4}{24}}{-4x^3 - 2x^4}.$

Divide numerator and denominator by $x^2$: $\frac{\text{Numerator}}{\text{Denominator}} \approx \frac{2 - \frac{7x}{6} + \frac{35x^2}{24}}{-4x - 2x^2}.$

As $x \to 0$, the dominant terms are: $\frac{2}{-4x} \to -\frac{1}{2x}.$

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