Question

Difficulty: HardLimits and Continuity of Functions
Evaluate the trigonometric limit:
limx0cos(3x)cos(x)x2\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}
What is the numerical value of this limit?

Answer: -4

Answer

The numerical value of the limit is -4.
Using either the sum-to-product identity cos(3x)cos(x)=2sin(2x)sin(x)\cos(3x) - \cos(x) = -2 \sin(2x) \sin(x) along with standard limits limx0sin(kx)x=k\lim_{x \to 0} \frac{\sin(kx)}{x} = k, or applying L'Hôpital's rule twice on the 00\frac{0}{0} form, yields the exact value 4-4.

Step-by-Step Solution

1
Check the form of the limit by direct substitution
Substituting x=0x = 0 gives cos(0)cos(0)02=110=00\frac{\cos(0) - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}, an indeterminate form.
Determines whether algebraic transformation or L'Hôpital's rule is required.
2
Transform the numerator using the sum-to-product formula
\cos(3x) - \cos(x) = -2 \sin\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = -2 \sin(2x) \sin(x)
Converts difference of cosines into product of sines to utilize standard trigonometric limits.
3
Rewrite the fractional expression and apply limit laws
\lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} = -2 \cdot \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right)
Splits x2x^2 into xxx \cdot x under each sine function.
4
Evaluate the individual standard limits
\lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1
Applies the known fundamental trigonometric limit rule limu0sin(au)u=a\lim_{u \to 0} \frac{\sin(au)}{u} = a.
5
Calculate the final product
-2 \times 2 \times 1 = -4
Combines all factors to reach the evaluated value.

Key Concept

Trigonometric Limits and Indeterminate Forms

Alternative Method

Alternatively, apply L'Hôpital's rule twice. First derivative of numerator over denominator yields limx03sin(3x)+sin(x)2x\lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x} (still 00\frac{0}{0}). Differentiating a second time yields limx09cos(3x)+cos(x)2=9(1)+12=82=4\lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4.
Estimated Time:1m 30s
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