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Squeeze theorem applied to x^2 cos(1/x^2) as x -> 0

Consider the function g(x)=x2cos⁡(1x2)g(x) = x^2 \cos\left(\dfrac{1}{x^2}\right), defined for all x≠0x \neq 0.

Use the Squeeze Theorem to evaluate lim⁡x→0x2cos⁡(1x2).\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right). Note that direct substitution fails here since cos⁡(1/x2)\cos(1/x^2) is undefined at x=0x=0 and oscillates wildly as x→0x \to 0. Start by bounding cos⁡(1/x2)\cos(1/x^2) between two constants, then multiply through by x2x^2 to construct upper and lower bounding functions whose limits you can compute directly. State the value of the limit.

Hint

Recall that −1≤cos⁡(θ)≤1-1 \le \cos(\theta) \le 1 for every real θ\theta. Substitute θ=1/x2\theta = 1/x^2 and multiply all parts of the inequality by x2x^2 (which is nonnegative, so the inequality directions don't flip) to get bounding functions g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x).

Step-by-step solution

Since −1≤cos⁡(1x2)≤1-1 \le \cos\left(\dfrac{1}{x^2}\right) \le 1 holds for all x≠0x \neq 0, multiply every part of this inequality by x2≥0x^2 \ge 0. This preserves the inequality directions and gives −x2≤x2cos⁡(1x2)≤x2-x^2 \le x^2\cos\left(\frac{1}{x^2}\right) \le x^2 for all x≠0x \neq 0.

Now compute the limits of the bounding functions: lim⁡x→0(−x2)=0\lim_{x\to 0} (-x^2) = 0 and lim⁡x→0x2=0\lim_{x\to 0} x^2 = 0, both by direct substitution since these are continuous polynomials. Since x2cos⁡(1/x2)x^2\cos(1/x^2) is squeezed between two functions that both tend to 00, the Squeeze Theorem gives lim⁡x→0x2cos⁡(1x2)=0.\lim_{x\to 0} x^2\cos\left(\frac{1}{x^2}\right) = 0.

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