A sine limit It can be shown that 1−x^2/ 6 ≤ sin x/ x ≤1, for x near 0.
Use these inequalities to evaluate lim x→0 sin x/ x.

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Start by understanding the given inequality: \(1 - \frac{x^2}{6} \leq \frac{\sin x}{x} \leq 1\) for \(x\) near 0. This inequality is derived from the Taylor series expansion of \(\sin x\).

Recognize that the limit we want to evaluate is \(\lim_{x \to 0} \frac{\sin x}{x}\).

Apply the Squeeze Theorem, which states that if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in some interval around \(c\) (except possibly at \(c\)), and \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\).

In this case, set \(f(x) = 1 - \frac{x^2}{6}\), \(g(x) = \frac{\sin x}{x}\), and \(h(x) = 1\). Notice that both \(f(x)\) and \(h(x)\) approach 1 as \(x\) approaches 0.

Conclude that by the Squeeze Theorem, \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). This is because both the lower bound \(1 - \frac{x^2}{6}\) and the upper bound 1 converge to 1 as \(x\) approaches 0.

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