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Squeeze theorem applied to x sin(1/x^2)

Consider the function h(x)=xsin⁡(1x2)h(x) = x \sin\left(\dfrac{1}{x^2}\right), which oscillates rapidly near x=0x = 0 but is multiplied by a factor that shrinks to 00.

Use the Squeeze Theorem to evaluate lim⁡x→0xsin⁡(1x2).\lim_{x \to 0} x \sin\left(\frac{1}{x^2}\right). Start by bounding sin⁡(1/x2)\sin(1/x^2) between −1-1 and 11, then multiply through by xx (being careful about the sign of xx) to build bounding functions whose limits as x→0x \to 0 you can compute directly. State the value of the limit.

Hint

Since −1≤sin⁡(1/x2)≤1-1 \le \sin(1/x^2) \le 1 for all x≠0x \neq 0, multiplying by ∣x∣|x| gives −∣x∣≤xsin⁡(1/x2)≤∣x∣-|x| \le x\sin(1/x^2) \le |x|. Find the limits of −∣x∣-|x| and ∣x∣|x| as x→0x \to 0 and apply the Squeeze Theorem.

Step-by-step solution

Since sine is always between −1-1 and 11, we know −1≤sin⁡(1/x2)≤1-1 \le \sin(1/x^2) \le 1 for all x≠0x \neq 0. Multiplying all sides of this inequality by ∣x∣≥0|x| \ge 0 (which does not flip the inequalities) gives −∣x∣≤xsin⁡(1x2)≤∣x∣.-|x| \le x\sin\left(\frac{1}{x^2}\right) \le |x|. This is valid because xsin⁡(1/x2)=∣x∣⋅sign(x)⋅sin⁡(1/x2)x \sin(1/x^2) = |x| \cdot \text{sign}(x) \cdot \sin(1/x^2) has absolute value at most ∣x∣|x|, so it lies between −∣x∣-|x| and ∣x∣|x|.

Now we compute 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, both by direct substitution since ∣x∣|x| is continuous. Since h(x)=xsin⁡(1/x2)h(x) = x\sin(1/x^2) is squeezed between two functions that both tend to 00 as x→0x \to 0, the Squeeze Theorem tells us lim⁡x→0xsin⁡(1x2)=0.\lim_{x\to 0} x\sin\left(\frac{1}{x^2}\right) = 0.

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