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Squeeze theorem applied to (sin x - x)/x^3

It is known that for all xx near 00 (with x≠0x \neq 0), the following inequality holds: 1−x26≤sin⁡(x)−xx3⋅(−1)≤11 - \frac{x^2}{6} \le \frac{\sin(x)-x}{x^3} \cdot (-1) \le 1

Actually, use the standard bounds derived from cos⁡(x)\cos(x) and sin⁡(x)\sin(x) near 00: for small x>0x > 0, one can show that −16≤sin⁡(x)−xx3≤−16cos⁡(x)-\frac{1}{6} \le \frac{\sin(x) - x}{x^3} \le -\frac{1}{6}\cos(x) This type of squeeze, built from the elementary bounds on sin⁡(x)\sin(x) and cos⁡(x)\cos(x) near 00, can be used to pin down the exact value of a limit that produces a 0/00/0 form under direct substitution.

Consider the function g(x)=sin⁡(x)−xx3.g(x) = \frac{\sin(x) - x}{x^3}.

Direct substitution at x=0x=0 gives the indeterminate form 0/00/0. Using the fact that near x=0x = 0 this expression can be trapped between two bounding functions built from cos⁡(x)\cos(x) and constants — specifically, for xx near 00 (and x≠0x \neq 0) one has −16≤sin⁡(x)−xx3≤−cos⁡(x)6,-\frac{1}{6} \le \frac{\sin(x)-x}{x^3} \le -\frac{\cos(x)}{6}, apply the Squeeze Theorem to evaluate lim⁡x→0sin⁡(x)−xx3.\lim_{x \to 0} \frac{\sin(x) - x}{x^3}.

What is the value of this limit?

Hint

Evaluate the limits of the lower bound −16-\tfrac{1}{6} and the upper bound −cos⁡(x)6-\tfrac{\cos(x)}{6} separately as x→0x \to 0, using lim⁡x→0cos⁡(x)=1\lim_{x\to 0}\cos(x) = 1. If both bounding limits agree, the Squeeze Theorem forces g(x)g(x) to have that same limit.

Step-by-step solution

Take the limit of the lower bounding function: lim⁡x→0(−16)=−16\lim_{x\to 0} \left(-\frac{1}{6}\right) = -\frac{1}{6}, since it is a constant function. This gives the lower limit needed for the squeeze.

Take the limit of the upper bounding function: lim⁡x→0(−cos⁡(x)6)=−cos⁡(0)6=−16\lim_{x\to 0} \left(-\frac{\cos(x)}{6}\right) = -\frac{\cos(0)}{6} = -\frac{1}{6}, using direct substitution since cos⁡(x)\cos(x) is continuous at x=0x=0.

Since both the lower bound −16-\frac{1}{6} and upper bound −cos⁡(x)6-\frac{\cos(x)}{6} approach the same value −16-\frac{1}{6} as x→0x \to 0, and g(x)g(x) is squeezed between them near 00, the Squeeze Theorem guarantees lim⁡x→0g(x)=−16\lim_{x\to 0} g(x) = -\frac{1}{6}.

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