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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4 Sullivan 132 Chapter 1 • Limits and Continuity Teaching Tip When studying the limit of functions as x approaches ∞ or −∞, it helps students to keep the following generalizations in mind: Infinity in the denominator: 2 y 4 2 4 2 4 2 x For example, the graph of f (x) = 1 in Figure 55 suggests that, as x becomes x unbounded in either the positive direction or the negative direction, the values f (x) get closer to 0. Table 14 illustrates this for a few numbers x. TABLE 14 x ±100 ±1000 ±10,000 ±100,000 x becomes unbounded f (x) = 1 x ±0.01 ±0.001 ±0.0001 ±0.00001 f (x) approaches 0 1. lim 1 = 0 x→∞ x 4 2. lim 1 = 0 x→−∞ x Infinity in the numerator: x 3. =∞ xlim →∞ 2 x 4. =−∞ xlim →−∞ 2 Square root and natural log of infinity: 5. ∞≈∞ lim x =∞ x→∞ Figure 55 f (x) = 1 x Since f can be made as close as we please to 0 by choosing x sufficiently large, we write 1 lim x→∞ x = 0 and say that the limit as x approaches infinity of f is equal to 0 . Similarly, as x becomes unbounded in the negative direction, the function f (x) = 1 can be made as close as we x please to 0, and we write lim x→−∞ 1 x = 0 1 Although we do not prove lim = 0 here, it can be proved using the ε-δ x→−∞ x Definition of a Limit at Infinity, and is left as an exercise in Section 1.6. The limit properties discussed in Section 1.2 hold for infinite limits. Although these properties are stated below for limits as x → ∞, they are also valid for limits as x → −∞. 6. ln( ∞) ≈∞ lim lnx =∞ x→∞ THEOREM Properties of Limits at Infinity If k is a real number, n ≥ 2 is an integer, and the functions f and g approach real numbers as x →∞, then the following properties are true: Alternate Example Finding Limits at Infinity Following are alternate ways to solve Example 5: a. lim 4 = 4 lim 1 x→−∞ 2 x x→−∞ 2 x = 4 lim 1 ⋅ lim 1 = 400 ⋅ ⋅ = 0 x→−∞ x x→−∞ x • lim A = A where A is a constant x→∞ • lim [kf(x)] = k lim f (x) x→∞ x→∞ • lim [ f (x) ± g(x)] = lim f (x) ± lim g(x) x→∞ x→∞ x→∞ [ ][ ] • lim [ f (x)g(x)] = lim f (x) lim g(x) x→∞ x→∞ x→∞ f (x) lim • f (x) lim x→∞ g(x) = x→∞ , provided lim g(x) = 0 lim g(x) x→∞ x→∞ [ ] n • lim [ f x→∞ (x)]n = lim f (x) x→∞ √ √ n • lim f (x) = n lim f (x), where f (x) >0 if n is even x→∞ x→∞ b. ⎛ 10 ⎞ ⎜ − ⎟ ⎝ ⎠ =− ⎛ lim 10 lim x→∞ x→∞⎜ x ⎝ =−10⋅ 0= 0 1 ⎞ ⎟ x ⎠ EXAMPLE 5 Find: (a) lim x→−∞ Finding Limits at Infinity ( 4 (b) lim − 10 ) √ x 2 x→∞ x (c) ( lim 2 + 3 ) x→∞ x ⎛ 3 ⎞ c. + ⎝ ⎜ ⎠ ⎟ = + ⎛ ⎞ lim 2 lim 2 3 lim 1 x→∞ x x→∞ x→∞⎝ ⎜ x⎠ ⎟ ⎛ ⎞ →∞ = 2+ 3 lim lim 1 x ⎝ ⎜ ⎠ ⎟ = 2 + 3 ⋅ 0 = 2 x→∞ lim x x→∞ 132 Chapter 1 • Limits and Continuity TE_Sullivan_Chapter01_PART II.indd 15 11/01/17 9:55 am
Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4 Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 133 Solution We use the properties of limits at infinity. 4 1 2 (a) lim x→−∞ x = 4 lim 1 2 = 4 lim = 4 · 0 = 0 2 x→−∞ x x→−∞ x ↑ 1 lim x→−∞ x = 0 (b) lim − 10 1 1 √ =−10 lim √ =−10 · lim x→∞ x x→∞ x x→∞ x 1 =−10 · lim x→∞ x =−10 ↑ · 0 = 0 1 lim x→∞ x = 0 (c) lim 2 + 3 = lim x→∞ x 2 + lim 3 x→∞ x→∞ x = 2 + 3 · lim 1 x→∞ x = 2 + 3 · 0 = 2 EXAMPLE 6 Find: (a) lim 3x 2 − 2x + 8 x→∞ x 2 + 1 Finding Limits at Infinity (b) lim 4x 2 − 5x x→−∞ x 3 + 1 ■ NOW WORK Problem 45. Solution (a) We find this limit by dividing each term of the numerator and the denominator by the term with the highest power of x that appears in the denominator, in this case, x 2 . Then 3x 2 − 2x + 8 lim x→∞ x 2 + 1 (b) = lim x→∞ ⏐ Divide the numerator and denominator by x 2 3x 2 − 2x + 8 x 2 x 2 + 1 x 2 3 − 2 = lim x + 8 x 2 x→∞ 1 + 1 = ⏐⏐⏐ x 2 Limit of a Quotient lim 3 − 2 x→∞ x + 8 x 1 2 + 1 x 2 lim x→∞ lim 3 − lim 2 x→∞ x→∞ = x + lim 8 1 3 − 2 lim x→∞ x 2 x→∞ x + 8 1 lim x→∞ x = lim 1 + lim 1 1 2 x→∞ x→∞ x 2 1 + lim x→∞ x = 3 − 0 + 0 = 3 ↑ 1 + 0 1 lim x = 0 x→∞ lim 4x 2 − 5x = lim x→−∞ x 3 + 1 ↑ x→−∞ Divide the numerator and denominator by x 3 TRM Section 1.5: Worksheet 2 This worksheet contains 6 questions in which the student finds the limit at infinity analytically. 4x 2 − 5x 4 x 3 x 3 = lim x − 5 x 2 + 1 x→−∞ 1 + 1 = x 3 x 3 lim x→−∞ lim x→−∞ 4 x − 5 x 2 2 1 + 3 x 3 = 0 1 = 0 ■ AP® CaLC skill builder for example 6 Finding Limits at Infinity x − x+ Find lim 2 3 10 . x→∞ 4 5− x Solution To find the limit at infinity, divide each term by the highest term in the denominator, that is, x 4 : 3 2x x 10 − + x − x+ lim 2 3 10 4 4 4 = lim x x x x→∞ 4 5 − x x→∞ 4 5 x − 4 4 x x = lim x→∞ 2 1 10 − + 3 4 x x x 5 −1 4 x lim 2 − lim 1 + lim 10 x→∞ →∞ →∞ = x x 3 x x 4 x lim 5 − lim 1 x→∞ 4 x x→∞ 0 = − 0 + 0 = 0 0−1 Alternate Example Finding Limits at Infinity Find x − x+ a. lim 3 2 2 8 x→∞ 2 x + 1 x − x b. xlim 4 2 5 →−∞ 3 x + 1 Solution Another way to evaluate the limits in Example 6 is to consider only the leading terms in the numerator and denominator. For instance, in (a), consider only 3x 2 in the numerator and x 2 in the denominator: x − x+ x lim 3 2 2 8 = lim 3 x→∞ 2 x + 1 x→∞ 2 x = lim 3= 3 x→∞ x − x x lim 4 2 5 = lim 4 2 = lim 4 = 0 →−∞ 3 x + 1 →−∞ 3 x →−∞ x x x x Your students may wonder why this method works. Go back to Example 6 and observe that every expression drops out except for the leading terms. This solution method is presented in Example 8 on page 135. 2 Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 133 TE_Sullivan_Chapter01_PART II.indd 16 11/01/17 9:55 am
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