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The given mathematical expression is:

\[
\lim_{{x \to 0}} \frac{e^x \sin x - x - x^2}{x^2 + x \log(1 - x)}
\]

To solve this limit, we will use **L'Hôpital's Rule**, since the limit initially results in the indeterminate form \( \frac{0}{0} \) as \( x \to 0 \). L'Hôpital's Rule states that:

\[
\lim_{{x \to a}} \frac{f(x)}{g(x)} = \lim_{{x \to a}} \frac{f'(x)}{g'(x)} \quad \text{if} \quad \frac{f(x)}{g(x)} \quad \text{gives the indeterminate form} \, \frac{0}{0} \text{ or } \frac{\infty}{\infty}.
\]

### Step 1: Check the limit when \( x \to 0 \)
We will first compute the individual limits of the numerator and the denominator when \( x \to 0 \).

#### Numerator: \( e^x \sin x - x - x^2 \)

- \( e^x \to 1 \) as \( x \to 0 \)
- \( \sin x \to x \) as \( x \to 0 \)

Therefore, \( e^x \sin x \to 1 \cdot 0 = 0 \). So:

\[
e^x \sin x - x - x^2 \to 0 - 0 - 0^2 = 0
\]

#### Denominator: \( x^2 + x \log(1 - x) \)

- \( x^2 \to 0 \)
- \( \log(1 - x) \) can be approximated using the Taylor series \( \log(1 - x) \approx -x \) for small \( x \), so \( x \log(1 - x) \to 0 \).

Thus, the denominator also approaches 0.

We now have the indeterminate form \( \frac{0}{0} \), which justifies the use of **L'Hôpital's Rule**.

### Step 2: Apply L'Hôpital's Rule

We will differentiate the numerator and the denominator with respect to \( x \).

#### Differentiate the numerator \( f(x) = e^x \sin x - x - x^2 \):
- \( \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x \)
- \( \frac{d}{dx}(-x) = -1 \)
- \( \frac{d}{dx}(-x^2) = -2x \)

So, the derivative of the numerator is:

\[
f'(x) = e^x (\sin x + \cos x) - 1 - 2x
\]

#### Differentiate the denominator \( g(x) = x^2 + x \log(1 - x) \):
- \( \frac{d}{dx}(x^2) = 2x \)
- \( \frac{d}{dx}(x \log(1 - x)) = \log(1 - x) + \frac{x}{1 - x} \)

So, the derivative of the denominator is:

\[
g'(x) = 2x + \log(1 - x) + \frac{x}{1 - x}
\]

### Step 3: Evaluate the new limit as \( x \to 0 \)

Now, substitute \( x = 0 \) into the derivatives:

- For the numerator: 
  \[
  f'(0) = e^0 (\sin 0 + \cos 0) - 1 - 2(0) = 1(0 + 1) - 1 - 0 = 0
  \]
  
- For the denominator:
  \[
  g'(0) = 2(0) + \log(1 - 0) + \frac{0}{1 - 0} = 0 + 0 + 0 = 0
  \]

Since we again obtain the indeterminate form \( \frac{0}{0} \), we need to apply **L'Hôpital's Rule** once more.

### Step 4: Apply L'Hôpital's Rule again

We differentiate the numerator and denominator again.

- The second derivative of the numerator \( f''(x) \) is:

\[
f''(x) = e^x (\sin x + 2\cos x) - 2
\]

Substituting \( x = 0 \):

\[
f''(0) = e^0 (\sin 0 + 2\cos 0) - 2 = 1(0 + 2(1)) - 2 = 2 - 2 = 0
\]

- The second derivative of the denominator \( g''(x) \) is more complex:

\[
g''(x) = 2 + \frac{d}{dx}\left(\log(1 - x) + \frac{x}{1 - x}\right)
\]

Substituting \( x = 0 \), we find that \( g''(0) = 2 \).

### Step 5: Conclusion

Finally, using L'Hôpital's Rule one last time, we find that:

\[
\lim_{{x \to 0}} \frac{f(x)}{g(x)} = \frac{f''(0)}{g''(0)} = \frac{0}{2} = 0
\]

Thus, the value of the limit is:

\[
\boxed{0}
\]

Would you like further details or have any questions? Here are some related questions for practice:

1. How do you find the Taylor series for \( e^x \sin x \) around \( x = 0 \)?
2. How would you solve this limit without using L'Hôpital's Rule?
3. What are the conditions for applying L'Hôpital's Rule?
4. How is the logarithmic approximation \( \log(1 - x) \approx -x \) derived?
5. What happens if the second derivatives also result in the indeterminate form?

**Tip:** L'Hôpital's Rule is very useful, but sometimes recognizing patterns in series expansions can make solving limits easier without resorting to differentiation.

Find the limit of (e^x * sin(x) - x - x^2) / (x^2 + x log(1 - x)) as x approaches 0.

Learn how to solve the limit lim (e^x sin x - x - x^2) / (x^2 + x log(1 - x)) as x approaches 0 using L'Hôpital's Rule. This step-by-step solution covers key calculus concepts like limits, indeterminate forms, and series expansions. Ideal for high school calculus students.

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