Question

Theory and ExamplesSuppose that g(x) ≤ f(x) ≤ h(x) for all x≠2 and suppose that lim x→2 g(x) = lim x→2 h(x) = −5. Can we conclude anything about the values of f, g, and h at x = 2? Could f(2) = 0? Could limx→2 f(x)=0? Give reasons for your answers.

Accepted Answer

First, let's understand the concept of the Squeeze Theorem, which is applicable here. The Squeeze Theorem states that if g(x) ≤ f(x) ≤ h(x) for all x in some interval around a point (except possibly at the point itself), and if the limits of g(x) and h(x) as x approaches that point are equal, then the limit of f(x) as x approaches that point is also equal to that common limit. Given that g(x) ≤ f(x) ≤ h(x) for all x ≠ 2 and both lim x→2 g(x) = −5 and lim x→2 h(x) = −5, by the Squeeze Theorem, we can conclude that lim x→2 f(x) = −5. Now, let's consider the value of f(2). The Squeeze Theorem does not provide information about the value of f at x = 2 itself; it only concerns the behavior of f(x) as x approaches 2. Therefore, f(2) could be any real number, including 0. Regarding whether lim x→2 f(x) could be 0, based on the Squeeze Theorem, we have already established that lim x→2 f(x) = −5. Therefore, lim x→2 f(x) cannot be 0. In summary, while the limit of f(x) as x approaches 2 is −5, the actual value of f at x = 2 is not determined by the Squeeze Theorem and could be any real number, including 0.

Theory and Examples

Suppose that g(x) ≤ f(x) ≤ h(x) for all x≠2 and suppose that lim x→2 g(x) = lim x→2 h(x) = −5. Can we conclude anything about the values of f, g, and h at x = 2? Could f(2) = 0? Could limx→2 f(x)=0? Give reasons for your answers.

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Key Concepts

Squeeze Theorem

Limits

Continuity