Question

Using the Sandwich Theorema. It can be shown that the inequalities 1 − x²/ 6 \u003c (x sin x) / (2−2cos x) \u003c 1 hold for all values of x close to zero (except for x = 0). What, if anything, does this tell you about limx→0 (x sin x) / (2 − 2cos x)?Give reasons for your answer.[Technology Exercise] b. Graph y = 1 − (x²/6), y=(x sinx)/(2 − 2cos x), and y = 1 together for −2 ≤ x ≤2. Comment on the behavior of the graphs as x→0.

Accepted Answer

Step 1: Understand the Sandwich Theorem, also known as the Squeeze Theorem. It states that if you have three functions f(x), g(x), and h(x) such that f(x) ≤ g(x) ≤ h(x) for all x in some interval around a point (except possibly at the point itself), and if limx→c f(x) = limx→c h(x) = L, then limx→c g(x) = L. Step 2: Apply the Sandwich Theorem to the given inequalities. We have 1 − x²/6 \u003c (x sin x) / (2 − 2cos x) \u003c 1 for values of x close to zero (except x = 0). We need to find the limits of the bounding functions as x approaches 0. Step 3: Calculate the limit of the lower bound function as x approaches 0. The function is 1 − x²/6. As x approaches 0, x² approaches 0, so the limit of 1 − x²/6 as x approaches 0 is 1. Step 4: Calculate the limit of the upper bound function as x approaches 0. The function is simply 1, which is constant. Therefore, the limit of 1 as x approaches 0 is also 1. Step 5: Conclude using the Sandwich Theorem. Since both the lower bound and upper bound functions approach the same limit of 1 as x approaches 0, by the Sandwich Theorem, the limit of (x sin x) / (2 − 2cos x) as x approaches 0 is also 1.

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Key Concepts

Sandwich Theorem

Limits

Graphical Analysis