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Squeeze theorem applied to a polynomial times an oscillating sine term

Consider the function f(x)=(x3−x2)sin⁡(1x)f(x) = (x^3 - x^2)\sin\left(\dfrac{1}{x}\right), defined for x≠0x \neq 0.

Use the Squeeze Theorem to evaluate lim⁡x→0(x3−x2)sin⁡(1x).\lim_{x \to 0} (x^3 - x^2)\sin\left(\frac{1}{x}\right). Since sin⁡(1/x)\sin(1/x) oscillates and is undefined at x=0x=0, direct substitution does not work. Instead, use the fact that −1≤sin⁡(1x)≤1-1 \le \sin\left(\dfrac{1}{x}\right) \le 1 for all x≠0x \neq 0 to bound f(x)f(x) above and below by two functions whose limits at x=0x=0 you can compute directly, then conclude the value of the original limit.

Hint

Start from −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 and multiply through by (x3−x2)(x^3 - x^2), being careful that the sign of x3−x2x^3-x^2 near 00 determines whether the inequality direction flips. Then take limits of the outer bounding functions as x→0x \to 0.

Step-by-step solution

Since −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for all x≠0x \ne 0, and near x=0x=0 we have x3−x2=x2(x−1)<0x^3 - x^2 = x^2(x-1) < 0 (because x−1x-1 is negative and x2x^2 is positive), multiplying the inequality by x3−x2x^3-x^2 reverses the inequality signs: (x3−x2)(1)≤(x3−x2)sin⁡(1/x)≤(x3−x2)(−1).(x^3-x^2)(1) \le (x^3-x^2)\sin(1/x) \le (x^3-x^2)(-1). This gives the bounds x3−x2≤f(x)≤−(x3−x2)=x2−x3x^3 - x^2 \le f(x) \le -(x^3-x^2) = x^2 - x^3 near x=0x=0.

Now compute the limits of the bounding functions using direct substitution: lim⁡x→0(x3−x2)=0−0=0\lim_{x\to 0}(x^3-x^2) = 0 - 0 = 0 and lim⁡x→0(x2−x3)=0−0=0\lim_{x\to 0}(x^2 - x^3) = 0 - 0 = 0. Since both bounding functions tend to 00 as x→0x \to 0, the Squeeze Theorem guarantees lim⁡x→0(x3−x2)sin⁡(1/x)=0.\lim_{x\to 0} (x^3-x^2)\sin(1/x) = 0.

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