Math Problem Statement

Lim x->0 cosx-cos2x/x²

Solution

We are tasked with evaluating the limit:

limx0cosxcos2xx2\lim_{x \to 0} \frac{\cos x - \cos 2x}{x^2}

Step-by-step solution:

  1. Apply L'Hopital's Rule: Since both the numerator and the denominator approach 0 as x0x \to 0 (an indeterminate form of 00\frac{0}{0}), we can apply L'Hopital's Rule. This requires taking the derivative of both the numerator and the denominator.

  2. First derivative of the numerator: The numerator is cosxcos2x\cos x - \cos 2x.

    • The derivative of cosx\cos x is sinx-\sin x.
    • The derivative of cos2x\cos 2x is 2sin2x-2\sin 2x (using the chain rule).

    So, the derivative of the numerator is: ddx[cosxcos2x]=sinx+2sin2x\frac{d}{dx} [\cos x - \cos 2x] = -\sin x + 2\sin 2x

  3. First derivative of the denominator: The denominator is x2x^2, and the derivative of x2x^2 is 2x2x.

  4. New limit expression: After applying L'Hopital's Rule once, we now evaluate the following limit: limx0sinx+2sin2x2x\lim_{x \to 0} \frac{-\sin x + 2\sin 2x}{2x} This is still an indeterminate form 00\frac{0}{0}, so we apply L'Hopital's Rule again.

  5. Second derivative of the numerator: The numerator is now sinx+2sin2x-\sin x + 2\sin 2x.

    • The derivative of sinx-\sin x is cosx-\cos x.
    • The derivative of 2sin2x2\sin 2x is 4cos2x4\cos 2x (using the chain rule again).

    So, the second derivative of the numerator is: ddx[sinx+2sin2x]=cosx+4cos2x\frac{d}{dx} [-\sin x + 2\sin 2x] = -\cos x + 4\cos 2x

  6. Second derivative of the denominator: The second derivative of 2x2x is just 2.

  7. Final limit expression: After applying L'Hopital's Rule again, we now evaluate: limx0cosx+4cos2x2\lim_{x \to 0} \frac{-\cos x + 4\cos 2x}{2} At x=0x = 0:

    • cos0=1\cos 0 = 1
    • cos2(0)=cos0=1\cos 2(0) = \cos 0 = 1

    Substituting these values into the limit: 1+4(1)2=1+42=32\frac{-1 + 4(1)}{2} = \frac{-1 + 4}{2} = \frac{3}{2}

Conclusion:

The value of the limit is: limx0cosxcos2xx2=32\lim_{x \to 0} \frac{\cos x - \cos 2x}{x^2} = \frac{3}{2}


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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
L'Hopital's Rule
Derivatives

Formulas

L'Hopital's Rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
Derivative of cos(x): -sin(x)
Chain rule for trigonometric functions

Theorems

L'Hopital's Rule
Chain Rule

Suitable Grade Level

Grades 11-12 and early university level