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Squeeze theorem applied to an oscillating trig limit

Consider the function f(x)=x4cos⁡(1x3)f(x) = x^4 \cos\left(\dfrac{1}{x^3}\right), which is undefined at x=0x=0 but oscillates rapidly as x→0x \to 0.

Use the Squeeze Theorem to evaluate lim⁡x→0x4cos⁡(1x3).\lim_{x \to 0} x^4 \cos\left(\frac{1}{x^3}\right). Clearly state the bounding functions you use, and explain why the Squeeze Theorem applies. What is the value of the limit?

Hint

Recall that cos⁡(θ)\cos(\theta) always satisfies −1≤cos⁡(θ)≤1-1 \le \cos(\theta) \le 1 for any real input θ\theta, even when θ=1/x3\theta = 1/x^3 blows up as x→0x \to 0. Multiply this inequality through by x4x^4 (which is nonnegative), and evaluate the limits of the resulting bounds as x→0x \to 0.

Step-by-step solution

Since cos⁡(θ)\cos(\theta) is always between −1-1 and 11 for any real number θ\theta, we have −1≤cos⁡(1/x3)≤1-1 \le \cos(1/x^3) \le 1 for all x≠0x \neq 0. Multiplying this inequality by x4≥0x^4 \ge 0 (which does not flip the inequality signs since x4x^4 is nonnegative) gives −x4≤x4cos⁡(1/x3)≤x4.-x^4 \le x^4\cos(1/x^3) \le x^4.

Now take the limit as x→0x \to 0 of the outer bounding functions: lim⁡x→0(−x4)=0\lim_{x\to 0}(-x^4) = 0 and lim⁡x→0x4=0\lim_{x\to 0} x^4 = 0 by direct substitution, since these are polynomials. Since f(x)=x4cos⁡(1/x3)f(x) = x^4\cos(1/x^3) is squeezed between two functions that both tend to 00, the Squeeze Theorem guarantees lim⁡x→0x4cos⁡(1/x3)=0.\lim_{x\to 0} x^4\cos(1/x^3) = 0.

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