Chap 3 Limits of Functions

Chap 3 Limits of Functions - FIRST YEAR CALCULUS W W L CHEN...

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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 3 LIMITS OF FUNCTIONS 3.1. Introduction We study the problem of the behaviour of a real valued function f (x) as the real variable x gets close to a given real number a, and begin by looking at a few simple examples. Example 3.1.1. Consider the function f (x) = x3 + x. Let us study its behaviour as x gets close to the real number 1, but is not equal to 1. We have the following numerical data: f (1.1) = 2.431, f (0.9) = 1.629, f (1.01) = 2.040301, f (0.99) = 1.960299, f (1.001) = 2.004003001, f (0.999) = 1.996002999. From this limited evidence, we suspect that f (x) is close to the value 2 when x is close to 1. Note here also that f (1) = 2. We would therefore like to say that x→1 lim f (x) = 2 = f (1). Example 3.1.2. Consider the function f (x) = (x3 − 1)/(x − 1). Let us study its behaviour as x gets close to the real number 1, but is not equal to 1. We have the following numerical data: f (1.1) = 3.31, f (0.9) = 2.71, f (1.01) = 3.0301, f (0.99) = 2.9701, f (1.001) = 3.003001, f (0.999) = 2.997001. From this limited evidence, we suspect that f (x) is close to the value 3 when x is close to 1. While the function f (x) is not defined at x = 1, we would nevertheless like to say that x→1 Chapter 3 : Limits of Functions lim f (x) = 3. page 1 of 13 First Year Calculus c W W L Chen, 1982, 2005 Example 3.1.3. Consider the function f (x) = x−1 sin x. Let us study its behaviour as x gets close to the real number 0, but is not equal to 0. From the graph, we suspect that f (x) is close to the value 1 when x is close to 0. While the function f (x) is not defined at x = 0, we would nevertheless like to say that lim f (x) = 1. x→0 Example 3.1.4. Consider the function f (x) = x−2 (1 − cos x). Let us study its behaviour as x gets close to the real number 0, but is not equal to 0. From the graph, we suspect that f (x) is close to the value 1 when x is close to 0. While the function 2 f (x) is not defined at x = 0, we would nevertheless like to say that lim f (x) = 1 . 2 x→0 Example 3.1.5. Consider the function f (x) = x sin(1/x). Let us study its behaviour as x gets close to the real number 0, but is not equal to 0. Chapter 3 : Limits of Functions page 2 of 13 First Year Calculus c W W L Chen, 1982, 2005 It appears that f (x) is close to the value 0 when x is close to 0. Let us look more closely. While the function f (x) is not defined at x = 0, we would nevertheless like to say that x→0 lim f (x) = 0. Example 3.1.6. Consider the function f (x) = x/|x|. Let us study its behaviour as x gets close to the real number 0, but is not equal to 0. Clearly f (x) = 1 when x > 0 and f (x) = −1 when x < 0. It follows that when x is close to 0, but not equal to 0, then f (x) is close to the value 1 or close to the value −1, depending on whether x is positive or negative. It is therefore clear that f (x) has no limit as x → 0. On the other hand, it is reasonable to say that f (x) is close to the value 1 when x > 0 is close to 0, and that f (x) is close to the value −1 when x < 0 is close to 0. In this case, we would like to say that x→1 lim f (x) does not exist, but also that x→0 x>0 lim f (x) = 1 and x→0 x<0 lim f (x) = −1. In order to formulate a proper definition for a limit, we need to study the differences |x − a| and |f (x) − L|, and find suitable ways to describe their smallness. In order to conclude that f (x) → L as x → a, we must therefore be able to convince ourselves that to make |f (x) − L| small, it is sufficient to make |x − a| small enough. Chapter 3 : Limits of Functions page 3 of 13 First Year Calculus c W W L Chen, 1982, 2005 Definition. We say that f (x) → L as x → a, or x→a lim f (x) = L, whenever 0 < |x − a| < δ . if, for every > 0, there exists δ > 0 such that |f (x) − L| < Remark. Note that we omit discussion of the case x = 1 in Example 3.1.2 and the case x = 0 in Examples 3.1.3–3.1.6. After all, we are only interested in those values of x which are close to a but not equal to a. The purpose of the restriction |x − a| > 0 is to omit discussion of the case when x = a. Example 3.1.7. Consider the function f (x) = 2x + 3. Let us study its behaviour as x → 1. Of course, we suspect that f (x) → 5 as x → 1. Here a = 1 and L = 5. We therefore need to study the differences |x − 1| and |f (x) − 5|. Let > 0 be chosen. Then |f (x) − 5| = |2x + 3 − 5| = |2x − 2| = 2|x − 1| < whenever |x − 1| < δ = /2. Example 3.1.8. Consider the function f (x) = x2 . Let us study its behaviour as x → 0. Of course, we suspect that f (x) → 0 as x → 0. Here a = 0 and L = 0. We therefore need to study the differences |x − 0| and |f (x) − 0|. Let > 0 be chosen. Then |f (x) − 0| = |x2 | < whenever |x − 0| = |x| < δ = √ . Example 3.1.9. Let us return to Example 3.1.1, and consider again the function f (x) = x3 + x when x → 1. We would like to show that f (x) → 2 as x → 1. Here a = 1 and L = 2. We therefore need to study the differences |x − 1| and |f (x) − 2|. Let > 0 be chosen. Then |f (x) − 2| = |x3 + x − 2| ≤ |x3 − 1| + |x − 1| = |x2 + x + 1||x − 1| + |x − 1|. Since we are only interested in those values of x close to 1, we shall lose nothing by considering only those values of x satisfying 0 < x < 2. Then |x2 + x + 1| = x2 + x + 1 < 7. It follows that if 0 < x < 2, then |f (x) − 2| < 8|x − 1| < if we have the additional restriction |x − 1| < /8. Note now that |x − 1| < 1 will guarantee 0 < x < 2. Hence |f (x) − 2| < can be guaranteed by |x − 1| < min{1, /8}. It follows that the requirements of the definition are satisfied if we take δ = min{1, /8}. Remark. The choice of δ is by no means unique. Suppose that in Example 3.1.9, we restrict our attention only to those values of x satisfying 0 < x < 1.5. Then |x2 + x + 1| = x2 + x + 1 < 5. It follows that if 0 < x < 1.5, then |f (x) − 2| < 6|x − 1| < if we have the additional restriction |x − 1| < /6. Note now that |x − 1| < 0.5 will guarantee 0 < x < 1.5. Hence |f (x) − 2| < can be guaranteed by |x − 1| < min{0.5, /6}. It follows that the requirements of the definition are satisfied also if we take δ = min{0.5, /6}. Indeed, in many situations, it will be very difficult, if not impossible, to obtain the best possible choice of δ . We are only interested in finding one value of δ that satisfies the requirements. Whether it is best possible or not is not important. Chapter 3 : Limits of Functions page 4 of 13 First Year Calculus c W W L Chen, 1982, 2005 3.2. Further Techniques The techniques of Examples 3.1.7–3.1.9 may be useful only in simple cases. If the given function is somewhat complicated, then the same approach will at best lead to a very complicated argument. An alternative is to seek ways to split the given function into “smaller” manageable parts. As an illustration, consider the function f (x) = x3 + x discussed in Example 3.1.9. We may choose to study the functions x3 and x separately, and note that the function x3 is the product of three copies of the simpler function x. The following result is called the Arithmetic of limits, comprising respectively the sum, product and quotient rules. See Section 3.6 for the proof. PROPOSITION 3A. Suppose that the functions f (x) → L and g (x) → M as x → a. Then (a) f (x) + g (x) → L + M as x → a; (b) f (x)g (x) → LM as x → a; and (c) if M = 0, then f (x)/g (x) → L/M as x → a. Remark. Note that for the quotient rule, we must impose the restriction that M = 0. Division by 0 is meaningless. Example 3.2.1. Consider the function h(x) = 2x3 + 5x + 3 x3 + 3x2 + 1 as x → 2. Clearly we have x2 → 4, x3 → 8. On the other hand, the constant function 2 → 2, so that the function 2x3 , being the product of the constant function 2 and the function x3 , satisfies 2x3 → 16 by the product rule. Similarly, we have 5x → 10 and 3x2 → 12. Naturally 3 → 3 and 1 → 1. It follows that as x → 2, we have h(x) = 2x3 + 5x + 3 16 + 10 + 3 29 → = . 3 + 3x2 + 1 x 8 + 12 + 1 21 Example 3.2.2. Consider the function h(x) = sin x + cos x sin x − 2 cos x as x → π/2. Here, we assume knowledge that sin x → 1 and cos x → 0 as x → π/2. Then clearly, as x → π/2, we have h(x) = sin x + cos x 1+0 → = 1. sin x − 2 cos x 1−0 A second alternative that we may pursue is to squeeze a given function between two known functions that have the same limit. As an illustration, consider the function f (x) = x sin x. Since −1 ≤ sin x ≤ 1 always, we have −|x| ≤ f (x) ≤ |x|. As x → 0, we clearly have |x| → 0. But then the function f (x) is squeezed between |x| and −|x| which both converge to 0. The following result is called the Squeezing principle. See Section 3.6 for the proof. PROPOSITION 3B. Suppose that g (x) ≤ f (x) ≤ h(x) for every x = a in some open interval that contains a. Suppose further that g (x) → L and h(x) → L as x → a. Then f (x) → L as x → a. Chapter 3 : Limits of Functions page 5 of 13 First Year Calculus c W W L Chen, 1982, 2005 Remark. It is crucial that squeezing occurs, in that g (x) and h(x) go to the same limit. To see that this is necessary, we use the well known result (see Problem 10) that the function f (x) = sin(1/x) does not approach a limit as x → 0. Clearly −1 ≤ f (x) ≤ 1, but squeezing does not occur. Example 3.2.3. We shall show that (1) x→0 lim sin x = 1. x To do this, we shall use some very simple geometric ideas to find two functions g (x) and h(x) to squeeze together. Suppose first of all that 0 < x < π/2. B D O x A C Let OAB be a right angled triangle formed by the points O(0, 0), A(cos x, 0) and B (cos x, sin x). Note then that the angle AOB has value x in radians. Note also that the points B and C (1, 0) both lie on the circle of radius 1 and centred at O. Finally, let D be the intersection point of the segment OB with the circle passing through A and centred at O. Suppose that we write α = area of circular segment OAD, β = area of triangle OAB, γ = area of circular segment OCB. Then clearly α < β < γ . On the other hand, simple calculation gives 2α = x cos2 x, 2β = sin x cos x and 2γ = x, so that (2) cos x < sin x 1 < . x cos x Note now that all the three terms in (2) remain unchanged if x is replaced by −x. It follows that (2) is valid for all x = 0 in the open interval (−π/2, π/2). Now take g (x) = cos x and h(x) = 1/ cos x. Then clearly g (x) → 1 and h(x) → 1 as x → 0. The assertion (1) now follows. Chapter 3 : Limits of Functions page 6 of 13 First Year Calculus c W W L Chen, 1982, 2005 3.3. One Sided Limits Recall Example 3.1.6, and consider also the following example. Example 3.3.1. Consider the function f (x) = x+2 x+3 if x > 3, if x ≤ 3. Then it is not difficult to see that as x → 3, the limit does not exist. On the other hand, it is easy to see that f (x) is close to the value 5 when x > 3 is close to 3, and that f (x) is close to the value 6 when x < 3 is close to 3. If we limit the approach to 3 to just from one side, then we can formulate one sided limits. Definition. We say that f (x) → L as x → a+, or x→a+ lim f (x) = L, whenever 0 < x − a < δ . In this case, L is if, for every > 0, there exists δ > 0 such that |f (x) − L| < called the right hand limit. Definition. We say that f (x) → L as x → a−, or x→a− lim f (x) = L, whenever 0 < a − x < δ . In this case, L is if, for every > 0, there exists δ > 0 such that |f (x) − L| < called the left hand limit. Example 3.3.2. Let us return to the function f (x) in Example 3.3.1. We have x→3− lim f (x) = 6 and x→3+ lim f (x) = 5. Example 3.3.3. Let us return to the function f (x) = x/|x| in Example 3.1.6. We have x→0− lim f (x) = −1 and x→0+ lim f (x) = 1. It is very easy to deduce the following result. PROPOSITION 3C. We have x→a lim f (x) = L if and only if x→a− lim f (x) = lim f (x) = L. x→a+ It is not difficult to formulate suitable analogues of the Arithmetic of limits and the Squeezing principle. Their precise statements are left as exercises. 3.4. Infinite Limits Consider the function f (x) = 1/x when x → 0. Although f (x) does not approach a finite limit, it is not difficult to accept that we can still say something about the behaviour of f (x) when x → 0, namely that f (x) gets rather large. Chapter 3 : Limits of Functions page 7 of 13 First Year Calculus c W W L Chen, 1982, 2005 Definition. We say that a function f (x) diverges to infinity, denoted by f (x) → ∞ as x → a, if, for every E > 0, there exists δ > 0 such that |f (x)| > E whenever 0 < |x − a| < δ . Example 3.4.1. Consider the function f (x) = 1/x. We suspect that f (x) → ∞ as x → 0. Here a = 0. Let E > 0 be chosen. Then |f (x)| = |1/x| = 1/|x| > E whenever |x − 0| = |x| < δ = 1/E . The following simple observation is useful. PROPOSITION 3D. The function f (x) → ∞ as x → a if and only if the function 1/f (x) → 0 as x → a. Example 3.4.2. Consider the function f (x) = 1/x sin x as x → 0. Let g (x) = 1/f (x) = x sin x. We shall first of all show that g (x) → 0 as x → 0. Let > 0 be given. Then |g (x) − 0| = |x sin x| ≤ |x| < whenever 0 < |x − 0| < δ if we choose δ = . It now follows from Proposition 3D that f (x) → ∞ as x → 0. Remark. Note that the Arithmetic of limits in Section 3.2 does not extend to infinite limits. Consider, for example, f (x) = 1/x and g (x) = −1/x. Then f (x) → ∞ and g (x) → ∞ as x → 0. Note, however, that f (x) + g (x) → 0 as x → 0. 3.5. Limits at Infinity We now study the behaviour of a function f (x) as x → +∞. The following definition is natural. Definition. We say that f (x) → L as x → +∞, or x→+∞ lim f (x) = L, whenever x > D. if, for every > 0, there exists D > 0 such that |f (x) − L| < Example 3.5.1. Consider the function f (x) = 1/x2 . Let us study its behaviour as x → +∞. Of course, we suspect that f (x) → 0 as x → +∞. Here L = 0. To prove this, let > 0 be chosen. Then |f (x) − 0| = |1/x2 | = 1/x2 < whenever x > D = 1 . We also study the behaviour of a function f (x) as x → −∞. Corresponding to the above, we have the following obvious analogue. Definition. We say that f (x) → L as x → −∞, or x→−∞ lim f (x) = L, whenever x < −D. page 8 of 13 if, for every > 0, there exists D > 0 such that |f (x) − L| < Chapter 3 : Limits of Functions First Year Calculus c W W L Chen, 1982, 2005 Example 3.5.2. Consider the function f (x) = 1 + x−1 sin x. Let us study its behaviour as x → −∞. Of course, we suspect that f (x) → 1 as x → −∞. After all, we have −1 ≤ sin x ≤ 1 always. Here L = 1. To prove this, let > 0 be chosen. Then, for x < 0, we have |f (x) − 1| = |x−1 sin x| ≤ |x−1 | = −x−1 < 1 whenever x < −D = − . [If you have difficulty following the calculation, note that if a < b and c < 0, then ac > bc. Check the calculation again.] Again, it is not difficult to formulate suitable analogues of the Arithmetic of limits and the Squeezing principle. Their precise statements are left as exercises. Finally, we have the following extra definitions which we seldom use. Definition. We say that f (x) → ∞ as x → +∞ if, for every E > 0, there exists D > 0 such that |f (x)| > E whenever x > D. Definition. We say that f (x) → ∞ as x → −∞ if, for every E > 0, there exists D > 0 such that |f (x)| > E whenever x < −D. 3.6. Further Discussion In this section, we shall establish Propositions 3A and 3B. Proof of Proposition 3A. (a) We shall use the inequality |(f (x) + g (x)) − (L + M )| ≤ |f (x) − L| + |g (x) − M |. Given any > 0, there exist δ1 , δ2 > 0 such that |f (x) − L| < /2 and |g (x) − M | < /2 whenever 0 < |x − a| < δ2 . whenever 0 < |x − a| < δ1 , Let δ = min{δ1 , δ2 } > 0. It follows that whenever 0 < |x − a| < δ , we have |(f (x) + g (x)) − (L + M )| ≤ |f (x) − L| + |g (x) − M | < . (b) We shall use the inequality |f (x)g (x) − LM | = |f (x)g (x) − f (x)M + f (x)M − LM | = |f (x)(g (x) − M ) + (f (x) − L)M | ≤ |f (x)||g (x) − M | + |M ||f (x) − L|. Since f (x) → L as x → a, there exists δ1 > 0 such that |f (x) − L| < 1 so that |f (x)| < |L| + 1 whenever 0 < |x − a| < δ1 . Chapter 3 : Limits of Functions page 9 of 13 whenever 0 < |x − a| < δ1 , First Year Calculus c W W L Chen, 1982, 2005 On the other hand, given any > 0, there exist δ2 , δ3 > 0 such that 2(|M | + 1) whenever 0 < |x − a| < δ2 , |f (x) − L| < and |g (x) − M | < 2(|L| + 1) whenever 0 < |x − a| < δ3 . Let δ = min{δ1 , δ2 , δ3 } > 0. It follows that whenever 0 < |x − a| < δ , we have |f (x)g (x) − LM | ≤ |f (x)||g (x) − M | + |M ||f (x) − L| < . (c) We shall first show that 1/g (x) → 1/M as x → a. To do this, we shall use the identity 1 1 |g (x) − M | − = . g (x) M |g (x)||M | Since M = 0 and g (x) → M as x → a, there exists δ1 > 0 such that |g (x) − M | < |M |/2 so that |g (x)| > |M |/2 On the other hand, given any whenever 0 < |x − a| < δ1 . whenever 0 < |x − a| < δ1 , > 0, there exists δ2 > 0 such that whenever 0 < |x − a| < δ2 . |g (x) − M | < M 2 /2 Let δ = min{δ1 , δ2 } > 0. It follows that whenever 0 < |x − a| < δ , we have 1 1 |g (x) − M | 2|g (x) − M | <. − = ≤ g (x) M |g (x)||M | |M |2 We now apply part (b) to f (x) and 1/g (x) to get the desired result. Proof of Proposition 3B. By Proposition 3A, we have h(x) − g (x) → 0 as x → a. We shall use the inequality |f (x) − L| = |(f (x) − g (x)) + (g (x) − L)| ≤ |f (x) − g (x)| + |g (x) − L| ≤ |h(x) − g (x)| + |g (x) − L|. Given any > 0, there exist δ1 , δ2 > 0 such that |h(x) − g (x)| < /2 and |g (x) − L| < /2 whenever 0 < |x − a| < δ2 . whenever 0 < |x − a| < δ1 , Let δ = min{δ1 , δ2 } > 0. It follows that whenever 0 < |x − a| < δ , we have |f (x) − L| ≤ |h(x) − g (x)| + |g (x) − L| < as required. Chapter 3 : Limits of Functions page 10 of 13 First Year Calculus c W W L Chen, 1982, 2005 Problems for Chapter 3 1. Use the definition of limit to prove each of the following: √ a) lim (4x + 5) = 13 b) lim x = 0 x→2 x→0 2. For each of the following functions, make a guess of the limit, then prove your assertion using the formal definition of a limit: a) f (x) = 3x + 5 as x → 2 b) f (x) = −4x + 5 as x → −1 c) f (x) = x2 as x → 0 d) f (x) = |x − 3| + 1 as x → 3 3. Consider f (x) = x2 . a) Find a δ > 0 so that |f (x) − 4| < 1/10 when |x − 2| < δ . b) Use the formal definition of a limit to prove that f (x) approaches 4 as x tends to 2. 4. a) Use the Arithmetic of limits to determine lim x2 + 3x + 2 . x→−1 2x2 − 8 b) By first factorizing the numerator and the denominator, determine lim all your steps carefully. x2 + 3x + 2 . Explain x→−2 2x2 − 8 √ 1 + 3x − 2 . x−1 5. Use the identity (a − b)(a + b) = a2 − b2 and the Arithmetic of limits to evaluate lim 6. Use the Arithmetic of limits in a suitable way to evaluate lim x→1 3x4 + 2x3 + 5x2 + 2 . x→+∞ 4x4 + 5x2 + x + 3 7. Use the Arithmetic of limits to determine each of the following, and explain each step carefully: 2x2 + 7x + 5 x2 − 5x + 1 a) lim b) lim 2 + 5x + 2 x→∞ 3x2 − 7x + 2 x→−1 3x 1 − cos 4x c) lim x2 − 3x + 1 − x d) lim x→+∞ x→0 x2 sin x [Hint: In part (d), the fact that lim = 1 may be useful. Do not try to use l’Hˆpital’s rule.] o x→0 x 8. Evaluate each of the following limits by using the Arithmetic of limits in a suitable way, and explain your steps carefully: x2 − 4 x2 − 4x + 3 x3 − 1 a) lim b) lim 2 c) lim 2 x→2 x 2 x→1 x − 5x + 4 x→1 x − 1 √ √ x+7−3 1 + 3x − 2 x1/3 − 1 d) lim f) lim e) lim √ x→2 x→1 x→1 x − 1 x−2 x+8−3 9. Use the Squeezing principle to find each of the following limits: 1 cos 3x a) lim b) lim x2 1 + cos x→∞ x→0 x x √ cos x2 d) lim e) lim x sin x cos x x→∞ x→0 x 1 x c) lim x2 sin x→0 10. Consider the function f (x) = sin(1/x), defined for x = 0. a) Show that for every δ > 0, there exist x1 , x2 ∈ (0, δ ) such that f (x1 ) = 1 and f (x2 ) = −1. b) Show that for every real number L ∈ R, we have |f (x1 ) − L| + |f (x2 ) − L| ≥ 2, where x1 and x2 are the solutions in (a). c) Show that for every real number L ∈ R and every δ > 0, there exists x0 ∈ (0, δ ) such that |f (x0 ) − L| ≥ 1. d) Explain why it is not true that f (x) → L as x → 0. Chapter 3 : Limits of Functions page 11 of 13 First Year Calculus c W W L Chen, 1982, 2005 11. The purpose of this problem is to prove that lim below: a) Let f (x) = (1 − cos x)/x. Convince yourself that f (x) = −f (−x) for every non-zero x ∈ R. b) Suppose first of all that 0 < x < π/2. Attempt to draw a diagram from the description below. Let OAB be a right angled triangle formed by the points O(0, 0), A(cos x, 0) and B (cos x, sin x), and note that the angle AOB has value x in radians. Note also that the points B and C (1, 0) both lie on the circle of radius 1 and centred at O. Using the fact that the length of the arc BC is greater than the length of the line segment BC , show that 0< 1 − cos x < cos x π−x 2 . 1 − cos x = 0. Follow carefully the steps indicated x→0 x c) Combining (a) and (b), deduce that for every real number x satisfying 0 < |x| < π/2, we have 0 < |f (x)| < cos π − |x| 2 . d) Prove that |f (x)| → 0 as x → 0. e) Use the definition of limits to show that the result follows from (d). 12. Suppose that f (x) → L as x → a. Prove that |f (x)| → |L| as x → a. 13. Prove that lim (x2 + x)1/2 − x1/2 1 =. 3 /2 x→0 2 x 14. Consider the function f (x) = (1 − cos x)/x. sin2 x . x(1 + cos x) b) Using the Arithmetic of limits and the results cos x → 1 and (sin x)/x → 1 as x → 0, show that f (x) → 0 as x → 0. You must explain each step carefully. 1 − cos x c) Evaluate lim . You must explain each step carefully. x→0 x2 a) Show that for every x ∈ R satisfying 0 < |x| < π/2, we have f (x) = cos x − 1 x→0 sin2 x 15. Find each of the following limits: sin(−5x) sin 3x a) lim b) lim x→0 x→0 tan(x/2) 7x c) lim 16. Evaluate each of the following limits by using the Arithmetic of limits in a suitable way, and explain your steps carefully: x+4 4x2 + x − 6 a) lim 2 b) lim x→+∞ x + x + 5 x→+∞ 5x2 − x + 10 3 x +1 c) lim 2 x2 + 4 − x d) lim x→+∞ x − 1 x→+∞ e) g) x→+∞ lim √ x2 + 4x + 3 − x x2 + 1 x f) h) x→−∞ lim x2 + 4x + 3 + x √ x2 + 1 x x→+∞ lim x→−∞ lim 17. Evaluate each of the following limits and explain your steps carefully: x2 − 9 x−2 1 a) lim b) lim x2 sin c) lim √ x→3 x − 3 x→0 x→2 x 2x2 + 1 − 3 2 x + 4x + 6 1 4 d) lim x2 cos e) lim f) lim x2 sin x→∞ x→0 x→0 x 4x2 + 3 x Chapter 3 : Limits of Functions page 12 of 13 First Year Calculus c W W L Chen, 1982, 2005 18. Evaluate each of the following limits if it exists: √ √ 9x2 + 4x + 5 9x2 + 4x + 5 a) lim b) lim x→+∞ x→−∞ x x √ 2 + x3 | sin x| x d) lim e) lim+ x→0 x x x→0 √ 2 + x3 x g) lim x→0 x c) lim 5x2 + x4 x→0 x √ 2 + x3 x f) lim− x x→0 √ 19. You are given that sin x → 0 and cos x → 1 as x → 0. Explain carefully how the sum, product and quotient rules of limits can be used to study the function x2 + sin x , cos x and calculate its limit as x → 0. 20. a) Show that lim sin 4x sin 2x sin 3x = 4 and lim = 6. x→0 sin x cos x x2 b) Use the results in part (a) and the Squeezing principle, or otherwise, to show that x→0 x→0 lim 3 sin 4x 2 sin 2x sin 3x − sin x cos x x2 sin(ecos x ) = 0. You must explain carefully each step of your argument. Chapter 3 : Limits of Functions page 13 of 13 ...
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This note was uploaded on 02/01/2009 for the course MATH 3423 taught by Professor Staff during the Spring '09 term at UCSD.

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