We are tasked with finding the limit:

$\lim_{x \to 0^+} \frac{\cos(x) - 2x}{2x \sin(x) + x^2 \cos(x)}$

Step 1: Direct Substitution

First, let's try direct substitution by plugging in $x = 0$ into both the numerator and the denominator:

Numerator:

$\cos(0) - 2(0) = 1 - 0 = 1$

Denominator:

$2(0) \sin(0) + (0)^2 \cos(0) = 0 + 0 = 0$

We have a form of $\frac{1}{0}$, which is undefined, suggesting that we may need to use L'Hopital's Rule, since we encounter a division by zero.

Step 2: Apply L'Hopital's Rule

L'Hopital's Rule is applicable for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. To apply it, we differentiate the numerator and the denominator separately with respect to $x$.

Derivative of the Numerator:

The numerator is $\cos(x) - 2x$, so the derivative is: $\frac{d}{dx}[\cos(x) - 2x] = -\sin(x) - 2$

Derivative of the Denominator:

The denominator is $2x \sin(x) + x^2 \cos(x)$. Using the product rule for both terms, the derivative is: $\frac{d}{dx}[2x \sin(x) + x^2 \cos(x)] = 2 \sin(x) + 2x \cos(x) + 2x \cos(x) - x^2 \sin(x)$ $= 2 \sin(x) + 4x \cos(x) - x^2 \sin(x)$

Now, we have a new expression for the limit:

$\lim_{x \to 0^+} \frac{-\sin(x) - 2}{2 \sin(x) + 4x \cos(x) - x^2 \sin(x)}$

Step 3: Evaluate the New Limit

Now, we substitute $x = 0$ into the new expressions.

Numerator:

$-\sin(0) - 2 = 0 - 2 = -2$

Denominator:

$2 \sin(0) + 4(0) \cos(0) - (0)^2 \sin(0) = 0 + 0 - 0 = 0$

We still get a form of $\frac{-2}{0}$, which suggests that the limit is infinite. Specifically, the sign of the denominator will tell us whether it approaches $+\infty$ or $-\infty$.

Step 4: Analyze the Denominator's Behavior

For small $x$, we can approximate the terms in the denominator as follows:

• $\sin(x) \approx x$ (for small $x$)
• $\cos(x) \approx 1$ (for small $x$)

So, for small $x$, the denominator becomes approximately:

$2x + 4x - 0 = 6x$

Thus, the denominator approaches $6x$ as $x \to 0^+$, and the numerator is $-2$. Therefore, the limit is:

$\lim_{x \to 0^+} \frac{-2}{6x} = -\infty$

Final Answer:

$\lim_{x \to 0^+} \frac{\cos(x) - 2x}{2x \sin(x) + x^2 \cos(x)} = -\infty$

Would you like to go over any part of the solution in more detail, or is there another question I can help with?

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Here are some related questions you might find interesting:

1. How does L'Hopital's Rule work for limits of the form $\frac{0}{0}$?
2. What happens when the denominator tends to 0 while the numerator tends to a non-zero value?
3. Can we use Taylor series expansions to approximate the behavior of the functions involved?
4. How does the approximation $\sin(x) \approx x$ and $\cos(x) \approx 1$ help in evaluating limits?
5. How do we handle limits involving trigonometric functions in general?

Tip: When dealing with limits that result in indeterminate forms, always check if L'Hopital's Rule can be applied or if approximations (like small angle approximations for sine and cosine) can simplify the process.