Question

Using the Sandwich Theorema. Suppose that the inequalities 1/2 − x² / 24 \u003c (1 − cos x)/ x² \u003c 1/2 hold for values of x close to zero, except for x = 0 itself. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about limx→0 (1 −cos x)/ x²?Give reasons for your answer.[Technology Exercise] b. Graph the equations y=(1/2) − (x²/24), y = (1 - cos x) / x², and y = 1/2 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 the limits of f(x) and h(x) as x approaches that point are equal, then the limit of g(x) as x approaches that point is the same. Step 2: Apply the Sandwich Theorem to the given inequalities. We have 1/2 - x²/24 \u003c (1 - cos x)/x² \u003c 1/2 for values of x close to zero. 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 f(x) = 1/2 - x²/24. As x approaches 0, x²/24 approaches 0, so the limit of f(x) is 1/2. Step 4: Calculate the limit of the upper bound function as x approaches 0. The function is h(x) = 1/2, which is constant. Therefore, the limit of h(x) as x approaches 0 is also 1/2. Step 5: Conclude using the Sandwich Theorem. Since both the lower bound and upper bound functions have the same limit of 1/2 as x approaches 0, by the Sandwich Theorem, the limit of (1 - cos x)/x² as x approaches 0 is also 1/2.

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Concetti chiave

Sandwich Theorem

Limit of a Function

Graphical Interpretation