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Evaluating a trigonometric limit using the known squeeze-theorem result

Recall the standard limit fact established via the Squeeze Theorem: lim⁡x→01−cos⁡(x)x2=12\displaystyle\lim_{x\to 0}\frac{1-\cos(x)}{x^2} = \frac{1}{2}.

Use the fact that lim⁡x→01−cos⁡(x)x2=12\displaystyle\lim_{x\to 0}\frac{1-\cos(x)}{x^2} = \frac{1}{2} to evaluate lim⁡x→01−cos⁡(3x)5x2.\lim_{x\to 0}\frac{1-\cos(3x)}{5x^2}.

Hint

Let u=3xu = 3x so that the numerator matches the standard form 1−cos⁡(u)1-\cos(u). Rewrite x2x^2 in terms of u2u^2, being careful to introduce the correct scaling factor, then apply the known limit lim⁡u→01−cos⁡uu2=12\lim_{u\to 0}\frac{1-\cos u}{u^2} = \frac12.

Step-by-step solution

Multiply and divide to rewrite the expression in terms of 3x3x: 1−cos⁡(3x)5x2=1−cos⁡(3x)(3x)2⋅9x25x2=1−cos⁡(3x)(3x)2⋅95.\frac{1-\cos(3x)}{5x^2} = \frac{1-\cos(3x)}{(3x)^2}\cdot\frac{9x^2}{5x^2} = \frac{1-\cos(3x)}{(3x)^2}\cdot\frac{9}{5}. This isolates the standard form 1−cos⁡(u)u2\frac{1-\cos(u)}{u^2} with u=3xu = 3x.

As x→0x\to 0, we also have u=3x→0u = 3x \to 0, so lim⁡x→01−cos⁡(3x)(3x)2=12\displaystyle\lim_{x\to 0}\frac{1-\cos(3x)}{(3x)^2} = \frac12 by the given squeeze-theorem result. Therefore the overall limit is 12⋅95=910\frac12\cdot\frac{9}{5} = \frac{9}{10}.

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