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Using known trigonometric limits and the product rule to evaluate a limit

Given the known limit lim⁡x→01−cos⁡(x)x=0\lim_{x \to 0} \frac{1-\cos(x)}{x} = 0 and the standard limit lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1, use limit laws to evaluate a related limit.

Consider the function v(x)=1−cos⁡(x)x2.v(x) = \frac{1-\cos(x)}{x^2}. By rewriting v(x)v(x) as a product of two factors whose limits as x→0x \to 0 are already known, use the product rule for limits to evaluate lim⁡x→01−cos⁡(x)x2.\lim_{x \to 0} \frac{1-\cos(x)}{x^2}.

Hint

Write 1−cos⁡(x)x2\dfrac{1-\cos(x)}{x^2} as 1−cos⁡(x)x⋅1x\dfrac{1-\cos(x)}{x} \cdot \dfrac{1}{x}... but that won't have a finite limit for each factor. Instead, try multiplying and dividing by (1+cos⁡(x))(1+\cos(x)) first, or split as 1−cos⁡xx⋅sin⁡xx⋅11+cos⁡x\dfrac{1-\cos x}{x}\cdot\dfrac{\sin x}{x}\cdot\dfrac{1}{1+\cos x} using 1−cos⁡2x=sin⁡2x1-\cos^2x=\sin^2x, then apply the product rule to the pieces with known limits.

Step-by-step solution

Multiply numerator and denominator by (1+cos⁡x)(1+\cos x): 1−cos⁡xx2⋅1+cos⁡x1+cos⁡x=1−cos⁡2xx2(1+cos⁡x)=sin⁡2xx2(1+cos⁡x).\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x} = \frac{1-\cos^2 x}{x^2(1+\cos x)} = \frac{\sin^2 x}{x^2(1+\cos x)}. This can be rewritten as a product: (sin⁡xx)⋅(sin⁡xx)⋅(11+cos⁡x).\left(\frac{\sin x}{x}\right)\cdot\left(\frac{\sin x}{x}\right)\cdot\left(\frac{1}{1+\cos x}\right).

Each factor has a known limit as x→0x \to 0: sin⁡xx→1\frac{\sin x}{x} \to 1 (twice), and 11+cos⁡x→11+1=12\frac{1}{1+\cos x} \to \frac{1}{1+1} = \frac{1}{2} by direct substitution since it's continuous at x=0x=0. By the product rule for limits, the limit of the product equals the product of the limits: 1⋅1⋅12=12.1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}.

Identifying the limit rule used to combine known factor limits

Hint
Step-by-step solution

The answer is the product rule for limits. Once the expression was rewritten as three factors each with a known limit, the product rule states that the limit of a product equals the product of the individual limits. This allowed multiplying one, one, and one half together to get the final answer of one half.

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