__limite functions.sect22-24
- 1. INFINITE LIMITS, LIMITS AT INFINITY, AND LIMIT RULES Sections 2.2, 2.4 & 2.5 September 5, 2013
- 2. INFINITE LIMITS Consider the function f(x) = 1 x . -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 As x → 0+ the value of f(x) = 1 x grows without bound (i.e. 1 x → ∞). As x → 0− the value of 1 x grows negatively without bound (i.e. 1 x → −∞). By definition a one or two-sided limit must be a real number (i.e. finite), so lim x→0+ 1 x and lim x→0− 1 x . do not exist.
- 3. INFINITE LIMITS However, it is convenient to describe the behavior of f(x) = 1 x from the left or right using “limit notation”. So we will write lim x→0+ 1 x = ∞ and lim x→0− 1 x = − ∞. This does NOT mean that the limit is ±∞, it is just a description of the behavior of our function f(x) near 0. We would say: “f(x) = 1 x approaches ∞ as x approaches 0 from the right.” “f(x) = 1 x approaches −∞ as x approaches 0 from the left.”
- 4. INFINITE LIMITS It may be the case that both one-sided limits are infinite and also the same. For example, lim x→0+ 1 x2 = ∞ = lim x→0− 1 x2 . In this case we will again abuse the notation and write lim x→0 1 x2 = ∞. We would say: “f(x) = 1 x2 approaches ∞ as x approaches 0.”
- 5. EXAMPLES f(x) = sec(x) = 1 cos(x) -5 -4 -3 -2 -1 0 1 2 3 4 5 g(x) = −1 (x − 7)2 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -4 -3 -2 -1 1 2 3 4 5 6 7 8 (1) lim x→ π 2 − f(x) = ∞ (2) lim x→ π 2 + f(x) = − ∞ (3) lim x→ π 2 f(x) = DNE (4) lim x→7− g(x) = ∞ (5) lim x→7+ g(x) = ∞ (6) lim x→7 g(x) = ∞
- 6. VERTICAL ASYMPTOTES Back to the function f(x) = 1 x . As noted before, the “one-sided limits” of f at 0 are both infinite. Consider the line x = 0. The graph of f gets arbitrarily arbitrarily close to the line x = 0, but it will never touch it. We call the line x = 0 a vertical asymptote of the function f(x) = 1 x .
- 7. VERTICAL ASYMPTOTES Definition A vertical line x = c is a vertical asymptote of the graph of a function f if either lim x→c+ f(x) = ± ∞ or lim x→c− f(x) = ± ∞.
- 8. VERTICAL ASYMPTOTES How do we find vertical asymptotes? Check wherever a denominator is zero. For example, f(x) = 1 x − 3 lim x→3− f(x) = −∞ lim x→3+ f(x) = ∞ Vert. Asym.: x = 3 f(x) = x2 − 1 x + 1 lim x→−1− f(x) = −2 lim x→−1+ f(x) = −2 Vert. Asym.: None f(x) = tan(x) For all odd integers n: lim x→ nπ 2 − f(x) = ∞ lim x→ nπ 2 + f(x) = −∞ Vert. Asym.: x = nπ/2 for all odd n
- 10. PRACTICE PROBLEMS Determine the following limits: (1) lim x→0 4 x2/5 (2) lim θ→(−π 2 )− sec(θ) (3) lim x→0+ x2 2 − 1 x Determine the equations of the asymptotes of the following functions: (1) f(x) = x2 − 1 2x + 4 (2) f(x) = sec(x) (3) f(x) = x3 + 1 x2
- 11. LIMIT RULES & THE SANDWICH THEOREM
- 12. LIMIT RULES Now that we’re (hopefully) familiar with limits, it would be nice if we had some rules for computing them more quickly. Let L, M c, and k be real numbers and lim x→c f(x) = L and lim x→c g(x) = M Sum & Difference Rules lim x→c (f(x) ± g(x)) = L ± M Product Rule lim x→c (f(x) · g(x)) = L · M
- 13. LIMIT RULES Let L, M c, and k be real numbers and lim x→c f(x) = L and lim x→c g(x) = M Quotient Rule lim x→c f(x) g(x) = L M if M = 0 Power Rule If r and s are integers with no common factor and s = 0 lim x→c (f(x))r/s = Lr/s If s is even, we also need L > 0.
- 14. LIMITS OF POLYNOMIALS For certain functions (continuous) finding the limit is easy. Polynomials are one of those types of functions. Polynomial Limit Rule Let P(x) be any polynomial, then lim x→c P(x) = P(c) By the quotient rule, if P(x) and Q(x) are polynomial with Q(c) = 0, we have lim x→c P(x) Q(x) = P(c) Q(c)
- 15. THE “ALMOST THE SAME” RULE The following rule is very useful. The “Almost the Same” Rule If f(x) = g(x) for all x = c in some interval open containing c, then lim x→c f(x) = lim x→c g(x) It’s not clear how one would use this, so let’s look at an example.
- 16. THE “ALMOST THE SAME” RULE Determine the following limit: lim t→−1 t2 + 3t + 2 t2 − t − 2 Since (−1)2 − (−1) − 2 = 0, we cannot use the quotient rule. However, notice that −1 is also a zero of the numerator: (−1)2 + 3(−1) + 2 = 0. This means that (t + 1) is a factor of both the numerator and the denominator. lim t→−1 t2 + 3t + 2 t2 − t − 2 = lim t→−1 (t + 1)(t + 2) (t + 1)(t − 2) = lim t→−1 (t + 2) (t − 2) = − 1 3
- 17. THE SANDWICH THEOREM Sometimes the limit rules aren’t enough to determine the limit of a function at a given point. The following theorem allows us to determine the limit of a function at such a point by “sandwiching” it between two other functions. The Sandwich Theorem Let f, g, and h be functions and c and L real numbers. If lim x→c g(x) = L = lim x→c h(x) and g(x) ≤ f(x) ≤ h(x) for all x in some open interval containing c, then lim x→c f(x) = L.
- 18. THE SANDWICH THEOREM This is a bit easier to see graphically: 66 Chapter 2: Limits and Continuity aid of som or throug useful. The San The follo Theorem of two ot tween the (Figure 2 x y 0 L c h f g FIGURE 2.12 The graph of ƒ is sandwiched between the graphs of g and h.
- 19. EXAMPLES We can use the sandwich theorem to prove two important limits: lim x→0 sin(x) = 0 and lim x→0 cos(x) = 1 In order to apply the sandwich theorem we need to find to functions to serve as the bread. In Chapter 1 the following two inequalities are proven (you wouldn’t be expected to know these): −|x| ≤ sin(x) ≤ |x| and 1 − |x| ≤ cos(x) ≤ 1. So, by the sandwich theorem we have lim x→0 −|x| = 0 = lim x→0 |x| =⇒ lim x→0 sin(x) = 0 lim x→0 1 − |x| = 1 = lim x→0 1 =⇒ lim x→0 cos(x) = 1.
- 20. EXAMPLES Use the sandwich theorem to determine the following limit: lim x→0 x2 sin(x) (Note that we could find this limit using the product rule.) In order to apply the sandwich theorem we need to find to functions to serve as the bread. Since −1 ≤ sin(x) ≤ 1 for all x = 0, we know that −x2 ≤ x2 sin(x) ≤ x2 So, by the sandwich theorem we have lim x→0 −x2 = 0 = lim x→0 x2 =⇒ lim x→0 x2 sin(x) = 0
- 21. EXAMPLES This is a bit easier to see graphically:
- 23. PRACTICE PROBLEMS Determine the following limits: (1) lim x→−2 x3 − 2x2 + 4x + 8 (2) lim t→6 8(t − 5)(t − 7) (3) lim y→2 y + 2 y2 + 5y + 6 (4) It can be shown that 1 − x2 6 ≤ x sin(x) 2 − 2 cos(x) ≤ 1. Use this fact to determine lim x→0 x sin(x) 2 − 2 cos(x)
- 25. LIMITS AT INFINITY Up to this point we’ve only considered the limit of a function at a real number c, that is, as x → c. We can also consider the limits of functions as x grows positively or negatively without bound, i.e. lim x→∞ f(x) and lim x→−∞ f(x) Limits at Infinity (intuitive) We say that a real number L is the limit of a function f as x approaches ∞ (resp., −∞), if the values of f(x) can be made arbitrary close to L by taking x to be sufficiently large (resp., sufficiently negative).
- 26. EXAMPLES Conisder the function f(x) = 1 x again.2.6 Limits Involving Infinity; Asymptotes of Graphs In this section we investigate the behavior of a fun pendent variable x becomes increasingly large, or of limit to infinite limits, which are not limits as b limit. Infinite limits provide useful symbols and l functions whose values become arbitrarily large in analyze the graphs of functions having horizontal o Finite Limits as The symbol for infinity does not represent a behavior of a function when the values in its dom For example, the function is defined positive and becomes increasingly large, be negative and its magnitude becomes increasingly summarize these observations by saying that or that 0 is a limit of at i precise definitions. ƒsxd = 1>xx : - q, ƒ 1>x ƒsxd = 1>x sqd x : —ˆ x y 0 1 –1 1–1 2 3 4 2 3 4 x 1 xy ϭ FIGURE 2.49 The graph of approaches 0 as or .x : - qx : q y = 1>x Notice that as x → ∞, the value of f(x) → 0. So, lim x→∞ 1 x = 0 Similarly, as x → −∞, the value of f(x) → 0. So, lim x→−∞ 1 x = 0
- 27. EXAMPLES For any integer n > 0, the graph of f(x) = 1 xn is given by: n even n odd So we see that lim x→±∞ 1 xn = 0 for all integers n ≥ 1.
- 28. EXAMPLES The limit of a function at ±∞ need not exist: f(x) = cos(x) lim x→±∞ cos(x) = DNE f(x) = x3 − x lim x→∞ x3 − x = ∞ lim x→−∞ x3 − x = −∞ f(x) = ex lim x→∞ ex = ∞ lim x→−∞ ex = 0
- 29. LIMIT RULES The limit rules also hold for limits at ±∞: Limit Rules Let L and M be real numbers and suppose lim x→±∞ f(x) = L and lim x→±∞ g(x) = M Then: lim x→±∞ (f(x) ± g(x)) = L ± M lim x→±∞ (f(x) · g(x)) = L · M lim x→±∞ (f(x)/g(x)) = L/M if M = 0 lim x→±∞ (f(x))r/s = Lr/s (r, s are integers with no common factors, and L > 0 if s is even)
- 30. THE SANDWICH THEOREM The sandwich theorem also hold for limits at ±∞: The Sandwich Theorem (for limits at ±∞) Let f, g, and h be functions and L real numbers, with lim x→∞ g(x) = L = lim x→∞ h(x) and g(x) ≤ f(x) ≤ h(x) for all sufficiently large x, then lim x→∞ f(x) = L. (The analogous statement holds for limits at −∞.)
- 31. EXAMPLE Use the sandwich theorem to determine the following limit lim x→±∞ sin(x) x We need to find functions that we can sandwich sin(x)/x between for sufficiently large and sufficiently negative x. Since −1 ≤ sin(x) ≤ 1, for all x = 0 we have − 1 x ≤ sin(x) x ≤ 1 x So, by the sandwich theorem lim x→±∞ sin(x) x = lim x→±∞ 1 x = 0.
- 33. LIMITS OF RATIONAL FUNCTIONS Even with our limit rules, determining limits at ±∞ can be difficult. However, for rational functions finding these limits is relatively easy. For example, suppose f(x) = x − 1 4x2 + 2 and we want to determine lim x→∞ f(x). First, we rewrite f(x) by dividing the numerator and the denominator by the term involving the largest power of x. In this case the term “4x2”: lim x→∞ x − 1 4x2 + 2 = lim x→∞ x 4x2 − 1 4x2 1 + 1 2x2 .
- 34. LIMITS OF RATIONAL FUNCTIONS Next we can apply the limit rules, specifically the quotient rule: lim x→∞ x − 1 4x2 + 2 = lim x→∞ x 4x2 − 1 4x2 1 + 1 2x2 = lim x→∞ x 4x2 − lim x→∞ 1 4x2 lim x→∞ 1 + lim x→∞ 1 2x2 = 0 − 0 1 + 0 = 0 Notice: The degree of the denominator was greater than that of the numerator and the limit was 0.
- 35. LIMITS OF RATIONAL FUNCTIONS Suppose f(x) = x2 − 1 4x2 + 2 . We rewrite f(x) by dividing the numerator and the denominator by the term involving the largest power of x and employ the quotient rule: lim x→∞ x2 − 1 4x2 + 2 = lim x→∞ 1 4 − lim x→∞ 1 4x2 lim x→∞ 1 + lim x→∞ 1 2x2 = 1 4 − 0 1 + 0 = 1 4 Notice: The degree of the denominator was equal to that of the numerator and the limit was a non-zero constant.
- 36. LIMITS OF RATIONAL FUNCTIONS Suppose f(x) = x3 − 1 4x2 + 2 . We rewrite f(x) by dividing the numerator and the denominator by the term involving the largest power of x and use the quotient rule: lim x→∞ x3 − 1 4x2 + 2 = lim x→∞ 1 − lim x→∞ 1 x3 lim x→∞ 4 x + lim x→∞ 2 x3 = 1 lim x→∞ 4 x + lim x→∞ 2 x3 = ∞ Notice: The degree of the denominator was less than that of the numerator and the limit was infinite.
- 37. LIMITS OF RATIONAL FUNCTIONS More generally we have: Theorem If f is a rational function, with f(x) = anxn + · · · + a1x + a0 bmxm + · · · + b1x + b0 . Then lim x→±∞ f(x) = lim x→±∞ anxn bmxm . Note that if m ≥ n lim x→±∞ f(x) = lim x→±∞ anxn bmxm = 0 m > n an bm m = n
- 38. EXAMPLES Determine the following limits: lim x→∞ 7x3 x3 − 3x2 + 6x = lim x→∞ 7x3 x3 = lim x→∞ 7 = 7 lim x→−∞ 3x + 7 x2 − 2 = lim x→∞ 3x x2 = lim x→∞ 3 x = 0 lim x→∞ 2x + x2 x − 4 = lim x→∞ x2 x = lim x→∞ x = ∞ lim x→−∞ −x4 x2 + 7x4 − 7x2 + 9 = lim x→∞ −x4 7x4 = lim x→∞ − 1 7 = − 1 7
- 39. HORIZONTAL ASYMPTOTES Definition We say that a function f has a horizontal asymptote of y = L if either lim x→−∞ f(x) = L or lim x→∞ f(x) = L. For example, as was just shown lim x→∞ 7x3 x3 − 3x2 + 6x = 7. So f(x) = 7x3 x3−3x2+6x has a horizontal asymptote of y = 7. -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
- 41. PRACTICE PROBLEMS For each of the functions below, determine the limit as x → ±∞ and the equation of any horizontal asymptotes: (1) f(x) = 3 − (2/x) 4 + (1/x2) (2) g(θ) = cos(θ) θ (3) h(y) = y + 2 y + 5 + 6y2 (4) q(x) = 7 + 3x2 x + 6