5.2 Limit Theorems

5.2. <a href="/tags/Limit_(mathematics)/" rel="tag">LIMIT</a> THEOREMS 1815.2 Limit Theorems5.2.1 Operations with Limits The same theorem we proved for sequences also hold for functions. We list the theorem, and leave its proof as an exercise.Theorem 5.2.1 Assuming that lim f (x) and lim g (x) exist, the following re- x a x a sults are true: → →1. lim (f (x) g (x)) = lim f (x) lim g (x) x a x a x a → ± → ± → 2. lim (f (x) g (x)) = lim f (x) lim g (x) x a x a x a → → → lim f (x) f (x) x a 3. lim = → as long as lim g (x) = 0 x a g (x) lim g (x) x a 6 → x a → → 4. lim f (x) = lim f (x) x a | | x a → → 5. If f (x) 0, then lim f (x) 0 x a ≥ → ≥ 6. If f (x) g (x) then lim f (x) lim g (x) x a x a ≥ → ≥ → 7. If f (x) 0, then lim f (x) = lim f (x) x a x a ≥ → → p q Remark 5.2.2 The above results also hold when the limits are taken as x . → ±∞ Remark 5.2.3 All the techniques learned in <a href="/tags/Calculus/" rel="tag">Calculus</a> can be used here. These techniques include factoring, multiplying by the conjugate.We look at a few examples to refresh the reader’smemory of some standard techniques. x2 6x + 8 Example 5.2.4 Find lim − x 4 x 4 Here, we note that both→ the numerator− and denominator are approaching 0 as x 4. Since both the numerator and denominator are polynomials, we know we→ can factor x 4. − x2 6x + 8 (x 4) (x 2) lim − = lim − − x 4 x 4 x 4 x 4 → → − = lim (x 2)− x 4 → − = 2 182 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION√x 1 Example 5.2.5 Find lim − x 1 x 1 Here, we note that both→ the numerator− and denominator are approaching 0 as x 1. The fraction also contains a radical. A standard technique is to try to multiply→ by the conjugate.√x 1 (√x 1) (√x + 1) lim − = lim − x 1 x 1 x 1 (x 1) (√x + 1) → − → − x 1 = lim − x 1 (x 1) (√x + 1) → − 1 = lim x 1 √x + 1 → 1 = 2 x2 + 5x + 1 Example 5.2.6 Find lim x 2x2 10 When taking the limit as→∞x −of a rational function, a standard technique is to factor the term of highest→ degree ±∞ from both the numerator and denominator.5 1 x2 1 + + x2 + 5x + 1 x x2 lim = lim x 2x2 10 x 10 →∞ − →∞ x2 2 − x2 5 1 10 As x , 1 + + 1 and 2 2, so → ∞ x x2 → − x2 → 5 1 x2 1 + + x x2 x2 lim = lim x 10 x 2x2 →∞ x2 2 →∞ − x2 1 = 2 Remark 5.2.7 The above example is a special case of a more general result we give below.Theorem 5.2.8 The limit as x of a rational function is the limit of the quotient of the terms of highest degree.→ ±∞ Proof. See problems.3x2 + 1 Example 5.2.9 Find lim x 4x3 + 2x + 1 →∞ 3x2 + 1 3x2 3 From the above theorem, we see that lim = lim = lim = x 4x3 + 2x + 1 x 4x3 x 4x 0. →∞ →∞ →∞ 5.2. LIMIT THEOREMS 1835.2.2 Elementary Theorems Theorems similar to those studied for sequences hold. We will leave the proof of most of these as an exercise.Theorem 5.2.10 If the <a href="/tags/Limit_of_a_function/" rel="tag">limit of a function</a> exists, then it is unique. Proof. See exercises at the end of this section.The next theorem relates the notion of limit of a function with the notion of <a href="/tags/Limit_of_a_sequence/" rel="tag">limit of a sequence</a>.Theorem 5.2.11 Suppose that lim f (x) = L and that xn is a sequence of x a → { } points such that lim xn = a, xn = a n. Put yn = f (xn). Then, lim yn = L n 6 ∀ n Proof. Let > →∞0 be given. Since lim f (x) = L, we can find →∞δ such that x a → 0 < x a < δ = f (x) L < . Since lim xn = a, we can find N such n | − | ⇒ | − | →∞ that n N = xn a < δ. But in this case, it follows that f (xn) L < . ≥ ⇒ | − | | − |The converse of this theorem is also true.Theorem 5.2.12 If f is defined in a deleted neighborhood of a such that f (xn) → L for every sequence xn such that xn a, then lim f (x) = L. { } → x a Proof. See problems at the end of this section. →Like for sequences, if a function has a limit at a point, then it is bounded. However, we need to be a little bit more careful here. If lim f (x) = L, then x a we are only given information about the behavior of f close→ to a. Therefore, we can only draw conclusions about what happens to f as long as x is close to a.Theorem 5.2.13 If lim f (x) = L, then there exists a deleted neighborhood of x a a in which f is bounded.→ Proof. We can find δ such that 0 < x a < δ = f (x) L < 1. By the | − | ⇒ | − | triangle inequality, we have for such x0s f (x) L f (x) L | | − | | ≤ || | − | || f (x) L ≤ | − | < 1It follows that f (x) < 1 + L | | | | Therefore, f is bounded by 1 + L in (a δ, a + δ). | | − Theorem 5.2.14 Suppose that f (x) g (x) h (x) in a deleted neighborhood of a and lim f (x) = lim h (x) = L then≤ lim g≤(x) = L. x a x a x a Proof. We→ do a direct→ proof here. In section→ 5.2.3 , we’ll see another proof. Suppose the above inequality holds in (a δ1, a + δ1) for some δ1 > 0. Let > 0 − 184 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION be given. Choose δ2 such that 0 < x a < δ2 = L < f (x) < +L. Choose | − | ⇒ − δ3 such that 0 < x a < δ3 = L < h (x) < + L. Let δ = min (δ1, δ2, δ3). Then, if 0 < x | a−<| δ, we have⇒ − | − | L < f (x) g (x) h (x) < + L − ≤ ≤ L < g (x) < + L ⇐⇒ − g (x) L < ⇐⇒ | − | Therefore, lim g (x) = L. x a → 1 Example 5.2.15 Show that lim x2 sin = 0. x 0 x → 1 We know that 1 sin 1 for every x R 0 . Hence, since − ≤ x ≤ ∈ \{ } 1 x2 0, we see that when x = 0 we have x2 x2 sin x2. Fur- ≥ 6 − ≤ x ≤ thermore, lim x2 = lim x2 = 0. By the squeeze theorem, it follows that x 0 − x 0 →1 → lim x2 sin �= 0. x 0 x → 5.2.3 Relationship Between the Limit of a Function and the Limit of a Sequence Theorems 5.2.11 and 5.2.12 can be summarized in the theorem below.Theorem 5.2.16 Let f be a function of one real variable defined in a deleted neighborhood of a real number a. The following conditions are equivalent.1. lim f (x) = L x a → 2. For every sequence xn such that xn = a and xn a we have lim f (xn) = { } 6 → n L. →∞Proof. We prove both directions.(1 = 2). We assume that lim f (x) = L. Let xn be a sequence such x a • ⇒ → { } that xn = a and xn a. Then, the conclusion follows from theorem 5.2.11. 6 → (2 = 1). This is theorem 5.2.12. • ⇒This theorem provides the link between the limit of a function and the limit of a sequence. Using this theorem, we can prove the theorems about the limit of a function by using their counterpart for sequences. We illustrate this with another version of the proof of the squeeze theorem. 5.2. LIMIT THEOREMS 185Theorem 5.2.17 Suppose that f (x) g (x) h (x) in a deleted neighborhood of a and lim f (x) = lim h (x) = L then≤ lim g≤(x) = L. x a x a x a → → → Proof. To show that lim g (x) = L, we need to show that if xn is any sequence x a → which converges to a, then g (xn) is a sequence which converges to L. Let xn be a sequence such that xn a, xn = a. Then, f (xn) L and h (xn) L. → 6 → → Since f (xn) g (xn) h (xn) we can apply the squeeze theorem for sequences ≤ ≤ to conclude that g (xn) L. → Theorem 5.2.16 also gives us a convenient way to show that a limit of a function does not exist. We summarize how in the following corollary.Corollary 5.2.18 Let f be a function of one real variable defined in a deleted neighborhood of a real number a. Then, lim f (x) does not exist if either of the x a following conditions holds: →1. There exist sequences (xn) and (yn) with xn = a and yn = a such that 6 6 lim xn = lim yn = a but lim f (xn) = lim f (yn). n n →∞ 6 →∞2. There exists a sequence (xn) with xn = a such that lim xn = a but the 6 sequence f (xn) diverges.We illustrate this corollary with an example.1 Example 5.2.19 Consider f : R 0 R given by f (x) = sin . Show { } → x lim f (x) does not exist. x 0 → 1 1 Consider the sequences xn = and yn = π . Then, clearly, lim xn = 2πn 2πn + 2 lim yn = 0. Yet, f (xn) = sin (2πn) = 0 thus lim f (xn) = 0 but f (yn) = n π →∞ sin 2πn + = 1 thus lim f (yn) = 1. 2 n →∞ 5.2.4 Exercises 1. Prove theorem 5.2.8.2. Prove theorem 5.2.10. Do both a direct proof and a proof using sequences.3. Prove theorem 5.2.1. Do both a direct proof and a proof using sequences.4. Evaluate the following limits: x2 4 (a) lim − x 2 x + 2 →− x3 27 (b) lim − x 3 x 3 → − 186 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION xn 1 (c) lim − where n is a positive integer. x 1 x 1 → xn− 1 (d) lim − x 1 xm 1 → − √x 2 2 (e) lim − − x 6 x 6 → − sin x 5. Assuming you know that lim = 1, compute the limits below: x 0 x → sin 2x (a) lim x 0 x → sin 3x (b) lim x 0 sin 5x → x (c) lim x 0 tan x → sin x (d) lim x 0 √x → (a + bx)(c + dx) √ac 6. Evaluate lim − x 0 x → p 7. Prove Theorem 5.2.12.8. Prove that if lim f (x) > 0 then there exists δ > 0 such that f (x) > 0 for x a every x (a →δ, a + δ). ∈ − 9. We say that a function f : I R where I R is Lipschitz providing there exists a constant K > 0 such→ that ⊆ f (x) f (y) K x y | − | ≤ | − | for x, y I. ∈ (a) Give an example of a Lipschitz function. You must show that the function in your example satisfies the required condition. (b) Prove rigorously that if f is Lipschitz, then lim f (x) = f (a). x a → 5.2. LIMIT THEOREMS 1875.2.5 Hints for the Exercises 1. Prove theorem 5.2.8. Hint: Write the expression of a typical rational function, then factor the term of highest degree from both the numerator and denominator.2. Prove theorem 5.2.10. Hint: The proof is similar to the same result for sequences.3. Prove theorem 5.2.1. Hint: Rewrite the result to prove using theorems 5.2.11 and 5.2.12, then use similar results for sequences.4. Evaluate the following limits: x2 4 (a) lim − x 2 x + 2 Hint:→− Use simple algebra to factor some terms. x3 27 (b) lim − x 3 x 3 Hint:→ Use− simple algebra to factor some terms. xn 1 (c) lim − where n is a positive integer. x 1 x 1 Hint:→ Use− simple algebra to factor some terms, remembering that n 2 n 1 x 1 = (x 1) 1 + x + x + ... + x − − − xn 1 (d) lim − � x 1 xm 1 Hint:→ Use− simple algebra to factor some terms, remembering that n 2 n 1 x 1 = (x 1) 1 + x + x + ... + x − − − √x 2 2 � (e) lim − − x 6 x 6 Hint:→ Use− the conjugate. sin x 5. Assuming you know that lim = 1, compute the limits below: x 0 x → sin 2x (a) lim x 0 x Hint:→ Use the fact that x 0 2x 0. → ⇐⇒ → sin 3x (b) lim x 0 sin 5x Hint:→ Remembering the previous problem, rewrite this problem so it sin x looks like so you can use the given result. x x (c) lim x 0 tan x Hint:→ Use the definition of tan x. 188 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION sin x (d) lim x 0 √x → Hint: In order to use the given result, you need to rewrite this so that there is an x in the denominator.(a + bx)(c + dx) √ac 6. Evaluate lim − x 0 x Hint: Use the→ p conjugate. 7. Prove Theorem 5.2.12. Hint: Using the definition of limits, try a proof by contradiction.8. Prove that if lim f (x) > 0 then there exists δ > 0 such that f (x) > 0 for x a every x (a →δ, a + δ). Hint: Using∈ − the definition of limits, pick an appropriate to have the desired result.9. We say that a function f : I R where I R is Lipschitz providing there exists a constant K > 0 such→ that ⊆ f (x) f (y) K x y | − | ≤ | − | for x, y I. ∈ (a) Give an example of a Lipschitz function. You must show that the function in your example satisfies the required condition. Hint: Try to understand the meaning of this condition. To help you f (x) f (y) do that, think of the geometric meaning of the quantity − . x y − (b) Prove rigorously that if f is Lipschitz, then lim f (x) = f (a). x a Hint: Simply use the definition of limit. → 5.3. MONOTONE FUNCTIONS 1895.3 Monotone FunctionsDefinition 5.3.1 (monotone increasing functions) A function f is said to be monotone increasing if whenever x1 > x2, then either f (x1) f (x2) or ≥ f (x1) > f (x2).We have the same terminology as for sequences. Like for sequences, functions which are non-decreasing (or non-increasing) have an important property.Theorem 5.3.2 Let f be a monotone function on the interval [a, b]. Then, the following limits exist:1. lim f (x) x b → − 2. lim f (x) x a+ → 3. lim f (x) for any interior point x0 + x xo →4. lim f (x) for any interior point x0 x xo− → Proof. Left as an exercise.Corollary 5.3.3 If f is non-decreasing on [a, b] then:1. f (a) lim f (x) ≤ x a+ → 2. lim f (x) f (b) x b ≤ → − 3. lim f (x) f (x0) lim f (x) for any interior point x0. + x xo− ≤ ≤ x xo → → 5.3.1 Exercises 1. Prove theorem 5.3.2.2. Prove corollary 5.3.3. Bibliography[B] Bartle, G. Robert, The elements of Real Analysis, Second Edition, John Wiley & Sons, 1976.[C1] Courant, R., Differential and Integral Calculus, Second Edition, Volume 1, Wiley, 1937. [C2] Courant, R., Differential and Integral Calculus, Second Edition, Volume 2, Wiley, 1937.[DS] Dangello, Frank & Seyfried Michael, Introductory Real Analysis, Houghton Miffl in, 2000. [F] Fulks, Watson, Advanced Calculus, an Introduction to Analysis, Third Edi- tion, John Wiley & Sons, 1978.[GN] Gaskill, F. Herbert & Narayanaswami P. P., Elements of Real Analysis, Prentice Hall, 1998. [LL] Lewin, J. & Lewin, M., An Introduction to <a href="/tags/Mathematical_analysis/" rel="tag">Mathematical Analysis</a>, Second Edition, McGraw-Hill, 1993. [MS] Stoll, Manfred, Introduction to Real Analysis, Second Edition, Addison- Wesley Higher Mathematics, 2001.501

5.2 Limit Theorems

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