Derivative of Cosine Function

Step 1: Verify the Indeterminate Form
Let's substitute $$x = 0$$ into the numerator and denominator to check the initial state of the limit:
* Numerator: $$e^{0 \cdot \cos 0} - 1 - 0 = e^0 - 1 = 1 - 1 = 0$$
* Denominator: $$\ln(1 + 0^2 - 0^3) = \ln(1) = 0$$

Since the expression yields the indeterminate form $$\frac{0}{0}$$, we can apply L'Hôpital's Rule.

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Step 2: Apply L'Hôpital's Rule (First Differentiation)
L'Hôpital's Rule states that if the limit is indeterminate of the form $$\frac{0}{0}$$, then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Let's find the derivative of the numerator using the chain rule and product rule:
$$\frac{d}{dx}\left(e^{x \cos x} - 1 - x\right) = e^{x \cos x} \cdot \frac{d}{dx}(x \cos x) - 1$$
$$= e^{x \cos x} \cdot (\cos x - x \sin x) - 1$$

Now let's find the derivative of the denominator:
$$\frac{d}{dx}\left(\ln(1 + x^2 - x^3)\right) = \frac{1}{1 + x^2 - x^3} \cdot (2x - 3x^2)$$

Substituting these derivatives back into the limit:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x) - 1}{\frac{2x - 3x^2}{1 + x^2 - x^3}}$$

To prevent the denominator algebra from becoming too messy, we can split out the term that does not approach zero:
$$\lim_{x \to 0} (1 + x^2 - x^3) = 1$$

Thus, we can rewrite the limit as:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x) - 1}{2x - 3x^2}$$

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Step 3: Verify the Indeterminate Form Again
Let's test $$x = 0$$ once more on our simplified expression:
* New Numerator: $$e^0(\cos 0 - 0) - 1 = 1(1) - 1 = 0$$
* New Denominator: $$2(0) - 3(0)^2 = 0$$

The limit is still in the indeterminate form $$\frac{0}{0}$$, so we apply L'Hôpital's Rule a second time.

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Step 4: Apply L'Hôpital's Rule (Second Differentiation)
Differentiate the numerator using the product rule:
$$\frac{d}{dx}\left[e^{x \cos x}(\cos x - x \sin x) - 1\right]$$
$$= \left[\frac{d}{dx}(e^{x \cos x})\right](\cos x - x \sin x) + e^{x \cos x}\left[\frac{d}{dx}(\cos x - x \sin x)\right]$$
$$= \left[e^{x \cos x}(\cos x - x \sin x)\right](\cos x - x \sin x) + e^{x \cos x}(-\sin x - \sin x - x \cos x)$$
$$= e^{x \cos x}(\cos x - x \sin x)^2 + e^{x \cos x}(-2\sin x - x \cos x)$$

Now differentiate the denominator:
$$\frac{d}{dx}(2x - 3x^2) = 2 - 6x$$

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Step 5: Substitute $$x = 0$$ to Find the Final Limit
Assemble the second derivatives back into the limit expression:
$$L = \lim_{x \to 0} \frac{e^{x \cos x}(\cos x - x \sin x)^2 + e^{x \cos x}(-2\sin x - x \cos x)}{2 - 6x}$$

Now substitute $$x = 0$$ directly:
$$L = \frac{e^0(\cos 0 - 0)^2 + e^0(-2\sin 0 - 0 \cdot \cos 0)}{2 - 6(0)}$$
$$L = \frac{1(1)^2 + 1(0)}{2}$$
$$L = \frac{1}{2}$$

Expand the functions in the numerator using Taylor series around zero:

$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$$

$$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$$

Substitute these series back into the numerator:

$$\text{Numerator} = a\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) + b\left(1 + x^2 + \frac{x^4}{2}\right) - 3$$

Group the terms by powers of $$x$$:

$$\text{Numerator} = (a + b - 3) + \left(b - \frac{a}{2}\right)x^2 + \left(\frac{a}{24} + \frac{b}{2}\right)x^4 + \dots$$

For the limit to be finite when dividing by $$x^4$$, the coefficients of $$x^0$$ and $$x^2$$ must be exactly zero:

1) $$a + b - 3 = 0 \implies a + b = 3$$

2) $$b - \frac{a}{2} = 0 \implies a = 2b$$

Substitute $$a = 2b$$ into the first condition:

$$2b + b = 3 \implies 3b = 3 \implies b = 1$$.