To evaluate the limit:

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

we will proceed step-by-step.

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Step 1: Substitution at $x = 0$

Direct substitution of $x = 0$ gives: $\frac{\cos(0) - 2(0)}{2(0) \sin(0) + (0)^2 \cos(0)} = \frac{1 - 0}{0 + 0} = \frac{1}{0}.$ This is indeterminate since the denominator approaches $0$. We proceed with L’Hôpital’s Rule as both the numerator and denominator tend to $0$ as $x \to 0^+$.

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Step 2: Apply L’Hôpital’s Rule

Differentiate the numerator and the denominator with respect to $x$:

Numerator:

$\frac{d}{dx}(\cos x - 2x) = -\sin x - 2.$

Denominator:

$\frac{d}{dx}(2x \sin x + x^2 \cos x) = 2 \sin x + 2x \cos x + 2x \cos x - x^2 \sin x.$ Simplify the derivative of the denominator: $\frac{d}{dx}(2x \sin x + x^2 \cos x) = 2 \sin x + 4x \cos x - x^2 \sin x.$

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Step 3: Rewrite the limit

The limit becomes: $\lim_{x \to 0^+} \frac{-\sin x - 2}{2 \sin x + 4x \cos x - x^2 \sin x}.$

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Step 4: Substitution at $x = 0$ again

Substituting $x = 0$:

• Numerator: $-\sin(0) - 2 = -2$,
• Denominator: $2 \sin(0) + 4(0) \cos(0) - (0)^2 \sin(0) = 0 + 0 - 0 = 0$.

Again, this is indeterminate, so we apply L’Hôpital’s Rule once more.

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Step 5: Apply L’Hôpital’s Rule again

Differentiate the numerator and denominator again:

Numerator:

$\frac{d}{dx}(-\sin x - 2) = -\cos x.$

Denominator:

Differentiate $2 \sin x + 4x \cos x - x^2 \sin x$: $\frac{d}{dx}(2 \sin x + 4x \cos x - x^2 \sin x) = 2 \cos x + 4 \cos x - 4x \sin x - 2x \sin x - x^2 \cos x.$ Simplify: $\frac{d}{dx}(2 \sin x + 4x \cos x - x^2 \sin x) = 6 \cos x - 6x \sin x - x^2 \cos x.$

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Step 6: Rewrite the limit

The limit becomes: $\lim_{x \to 0^+} \frac{-\cos x}{6 \cos x - 6x \sin x - x^2 \cos x}.$

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Step 7: Substitution at $x = 0$

Substitute $x = 0$:

• Numerator: $-\cos(0) = -1$,
• Denominator: $6 \cos(0) - 6(0) \sin(0) - (0)^2 \cos(0) = 6 - 0 - 0 = 6$.

Thus, the limit is: $\lim_{x \to 0^+} \frac{-\cos x}{6 \cos x - 6x \sin x - x^2 \cos x} = \frac{-1}{6}.$

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Final Answer:

$\boxed{-\frac{1}{6}}$

Let me know if you would like further clarification or additional examples. 😊

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Five Related Questions:

1. How is L’Hôpital’s Rule applied to solve indeterminate forms?
2. What are common indeterminate forms encountered in calculus?
3. Can you evaluate $\lim_{x \to 0^+} \frac{\sin x}{x^2 + \cos x - 1}$?
4. How do trigonometric approximations (e.g., $\sin x \approx x$ for small $x$) help in solving limits?
5. What are alternative methods to solve limits when L’Hôpital’s Rule fails?

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Tip:

When using L’Hôpital’s Rule, always check that the function is in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying the rule.