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Squeeze theorem applied to x cos(1/x) as x → 0

Consider the function g(x)=xcos⁡(1x)g(x) = x\cos\left(\dfrac{1}{x}\right), which is undefined at x=0x = 0 but whose limit as x→0x \to 0 can still be determined.

Use the Squeeze Theorem to evaluate lim⁡x→0xcos⁡(1x).\lim_{x \to 0} x\cos\left(\frac{1}{x}\right). Start by using the fact that cos⁡(1/x)\cos(1/x) is always between −1-1 and 11 for x≠0x \neq 0 to build a pair of bounding functions for g(x)=xcos⁡(1/x)g(x) = x\cos(1/x), then take the limit of each bound as x→0x \to 0. State the value of the limit.

Hint

Since −1≤cos⁡(1/x)≤1-1 \le \cos(1/x) \le 1 for all x≠0x \neq 0, multiply this inequality by xx (being careful that multiplying by a negative xx reverses the inequality) to get bounds of the form −∣x∣≤xcos⁡(1/x)≤∣x∣-|x| \le x\cos(1/x) \le |x|. Then find the limits of the outer bounding functions as x→0x \to 0.

Step-by-step solution

Since −1≤cos⁡(1/x)≤1-1 \le \cos(1/x) \le 1 holds for every x≠0x \neq 0, multiplying through by ∣x∣|x| (which is nonnegative, so the inequality direction is preserved) gives −∣x∣≤xcos⁡(1/x)≤∣x∣-|x| \le x\cos(1/x) \le |x|. This is valid because xcos⁡(1/x)x\cos(1/x) has the same sign behavior as xx times a bounded quantity, so its absolute value never exceeds ∣x∣|x|.

Now consider the limits of the bounding functions: lim⁡x→0(−∣x∣)=0\lim_{x\to 0}(-|x|) = 0 and lim⁡x→0∣x∣=0\lim_{x\to 0}|x| = 0. Since −∣x∣≤xcos⁡(1/x)≤∣x∣-|x| \le x\cos(1/x) \le |x| near x=0x=0 (excluding x=0x=0 itself) and both outer limits equal 00, the Squeeze Theorem guarantees that lim⁡x→0xcos⁡(1/x)=0\lim_{x\to 0} x\cos(1/x) = 0 as well.

Squeeze theorem applied to x cos(1/x) as x → 0 — solution graph

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