(II) For each real number ǫ > 0 there exists a real number δ(ǫ) such that 0 < δ(ǫ) δ 0 and

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1 9 44 One-sided its Definition 44 A function f has the it L R as x approaches a real number a from the left if the following two conditions are satisfied: (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a δ 0, a (II For each real number ǫ > 0 there exists a real number δ(ǫ such that 0 < δ(ǫ δ 0 0 < a x < δ(ǫ f(x L < ǫ If the conditions (I (II in Definition 44 are satisfied we write f(x = L x a Definition 44 A function f has the it L R as x approaches a real number a from the right if the following two conditions are satisfied: (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a, a+δ 0 (II For each real number ǫ > 0 there exists a real number δ(ǫ such that 0 < δ(ǫ δ 0 0 < x a < δ(ǫ f(x L < ǫ If the conditions (I (II in Definition 44 are satisfied we write f(x = L x a Definition 443 A function f has the it + as x approaches a real number a from the left if the following two conditions are satisfied: (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a δ 0, a (II For each real number M > 0 there exists a real number δ(m such that 0 < δ(m δ 0 0 < a x < δ(m f(x > M If the conditions (I (II in Definition 443 are satisfied we write f(x = + x a Definition 444 A function f has the it + as x approaches a real number a from the right if the following two conditions are satisfied: (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a, a+δ 0 (II For each real number M > 0 there exists a real number δ(m such that 0 < δ(m δ 0 0 < x a < δ(m f(x > M If the conditions (I (II in Definition 444 are satisfied we write f(x = + x a Definition 445 A function f has the it as x approaches a real number a from the left if the following two conditions are satisfied:

2 0 (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a δ 0, a (II For each real number M < 0 there exists a real number δ(m such that 0 < δ(m δ 0 0 < a x < δ(m f(x < M If the conditions (I (II in Definition 445 are satisfied we write f(x = x a Definition 446 A function f has the it as x approaches a real number a from the right if the following two conditions are satisfied: (I There exists a real number δ 0 > 0 such that f(x is defined for each x in the set ( a, a+δ 0 (II For each real number M < 0 there exists a real number δ(m such that 0 < δ(m δ 0 0 < x a < δ(m f(x < M If the conditions (I (II in Definition 446 are satisfied we write f(x = x a Exercise 447 Find each of the following its Prove your claims using the appropriate definition 3x 5 3x 5 x 3 (a (b (c x 5 x 0x + 5 x 5 x 0x + 5 x x(x (d x 0 ( x x (e 6 (f x 5 5 x x 5 5 x (g x + 3 x 3 x 9 (h x x 3 x 9 (i ( x x x 0 (j x 3 x x (k (x 3 x x + (l ( x x x +

3 5 Continuous Functions 5 Definition Examples All this work about its will now pay off since we shall be able to give mathematically rigorous definition of a continuous function Definition 5 Let f be a real valued function of a real variable let a be a real number The function f is continuous at a if the following two conditions are satisfied: (i The function f is defined at a, that is f(a is defined (ii f(x = f(a To underst Definition 5 the reader has to underst the concept of it Sometimes it is useful to state the definition of continuity directly, without appealing to the concept of it Definition 5 Let f be a real valued function of a real variable let a be a real number The function f is continuous at a if the following two conditions are satisfied: (I There exists a δ 0 > 0 such that f(x is defined for all x (a δ 0, a + δ 0 (II For each ǫ > 0 there exists δ(ǫ such that 0 < δ(ǫ δ 0 such that x a < δ(ǫ f(x f(a < ǫ Definition 5 is called ǫ-δ definition of continuity Example 53 Let c be a real number define f(x = c for all x R Use Definition 5 to prove that f is continuous at an arbitrary real number a Example 54 Let f(x = x for all x R Use Definition 5 to prove that f is continuous at an arbitrary real number a Example 55 Use ǫ-δ definition of continuity, that is Definition 5, to prove that the function f(x = /x is continuous on the interval (0, + Solution Let a (0, +, that is let a be an arbitrary positive number Chose δ 0 = a/ Since a > 0, we conclude that a/ > 0 f(x = /x is defined for all x ( a/, 3a/ Let ǫ > 0 be arbitrary Now we have to solve x a < ǫ for x a First simplify the expression, using the fact that x > 0 a > 0 rules for the absolute value: x a = a x a x x a xa = = x a xa

4 To get a larger expression which will be easy to solve we replace x in the denominator by the smallest possible value for x That value is a a/ = a/ This gives me my BIN: x x a x a a = xa a a = x a a Thus my BIN is x x a a valid for all x ( a/, 3a/ a x a Solving < ǫ for x a is easy: The solution is x a < a ǫ/ Now we define a { a ǫ δ(ǫ = min, a } It remains to prove the implication x a < δ(ǫ x a < ǫ This should be easy, using the BIN Example 56 Use ǫ-δ definition of continuity, that is Definition 5, to prove that the function x x is continuous on the interval (0, + Solution Let a (0, + Chose δ 0 = a Since a > 0, as before we conclude that a > 0 the function x x is defined for all x (a/, 3a/ Let ǫ > 0 be arbitrary Now we have to solve x a < ǫ for x a First simplify algebraically the expression, using the fact that x > 0 a > 0 rules for the absolute value x a ( = x a = ( x + a x a = x a x + a x + a x a = x + x a x a = a x + a a Thus my BIN is: Solving x a x a a, valid for x > 0 x a a < ǫ for x a is easy: The solution is x a < a ǫ Now we define { a a δ(ǫ = min ǫ, It remains to prove the implication x a < min should be easy, using the BIN } { a ǫ, a } x a < ǫ This Example 57 Let f(x = for all x R Use ǫ-δ definition to prove that f is continuous x + at an arbitrary a R

5 3 Theorem 58 (Algebra of Continuous Functions Let f g be functions let a be a real number Assume that f g are continuous at the point a (a If h = f + g, then h is continuous at a (b If h = fg, then h is continuous at a (c If h = f g g(a 0, then h is continuous at a Example 59 Let a, b, c be any real numbers Let f(x = ax + bx + c for all x R Let v be an arbitrary real number Prove that f is continuous at v Example 50 Let f(x = for all x R Use algebra of continuous functions to prove x + that f is continuous on R Example 5 Let f(x = sin x for all x R Prove that f is continuous at an arbitrary real number a Solution The proof consists of two steps: ( Use the definition of it an inequalities that we proved for sin cos to prove the following two its sin(x a = 0, cos(x a = ( Use the addition formula for sin x = sin(x a + a the algebra of its to complete the proof Example 5 Let f(x = cosx for all x R Prove that f is continuous at an arbitrary real number a Example 53 Let f(x = tanx for all π < x < π Prove that f is continuous at an arbitrary real number a such that π < a < π Solution Use the algebra of continuous functions Example 54 Let f(x = lnx for all x (0, + Prove that f is continuous at an arbitrary real number a Solution Use the definition of ln to derive the squeeze for ln: ln x x, 0 < x < + x ( x Use the above squeeze to prove that for arbitrary a > 0 we have ln a rule for logarithms lnuv = ln u + ln v = 0 Now use the

6 4 Example 55 Let f(x = e x for all x R Prove that f is continuous at an arbitrary real number a Solution Use the fact that x e x is the inverse of the logarithm function to derive the squeeze for it: + x e x, < x < x Get the rest of the proof as an exercise Theorem 56 Let f g be functions let a be a real number Assume that g is continuous at a that f is continuous at g(a If h = f g then h is continuous at a 6 New Limits from Old 6 Squeeze theorems In this section in Section 63 we establish general properties of its which are based on Definition 4 These properties are stated as theorems Establishing theorems of this kind involves a major step forward in sophistication Up to this point we have been trying to show that its exist directly from the definition Now for the first time we are going to assume that some it exists (I refer to this in class as a green it try to make use of this information to establish the existence of some other it (I refer to this in class as a red it Remember that to establish the existence of a it, we had to come up with a procedure for finding δ(ǫ that will work for any ǫ > 0 that is given If we assume the existence of a it, then we are assuming the existence of such a procedure, though we may not know explicitly what it is I refer to this as a green δ(ǫ It is this procedure we will need to use in order to construct a new procedure for the it whose existence we are trying to establish I refer to this as a red δ(ǫ We start by considering squeeze theorems that resemble the role of BIN in previous sections The following theorem is the Swich Squeeze Theorem Theorem 6 Let f, g h be given functions let a L be real numbers Suppose that the following three conditions are satisfied ( f(x = L, ( h(x = L, (3 There exists η 0 > 0 such that f(x, g(x h(x are defined for all x ( a η 0, a ( a, a+η 0 f(x g(x h(x for all x ( a η 0, a ( a, a + η 0 Then g(x = L

7 5 Proof Here we have three functions three definitions of its, one for each function Therefore we have to deal with three δ-s We shall give them appropriate names that will distinguish them from each other Let us name them δ f, δ g δ h In the theorem it is assumed that f(x = L This means that we are given the fact that for each ǫ > 0 there exists δ f (ǫ > 0 (that is, we are given a function δ f (ǫ such that 0 < x a < δ f (ǫ f(x L < ǫ (6 In class I refer to these as a green δ f ( a green implication Since the theorem assumes that h(x = L, we are also given that for each ǫ > 0 there exists δ h (ǫ > 0 such that 0 < x a < δ h (ǫ h(x L < ǫ (6 Again we refer to these as a green δ h ( a green implication We need to prove that g(x = L Therefore, following Definition 4, we have to show that the following conditions are satisfied: (I There exists a real number δ 0,g > 0 such that g(x is defined for each x in the set ( a δ 0,g, a ( a, a + δ 0,g (II For each real number ǫ > 0 there exists a real number δ g (ǫ such that 0 < δ g (ǫ δ 0,g such that 0 < x a < δ g (ǫ g(x L < ǫ (63 Since we have to produce δ 0,g, δ g (ǫ we have to prove the last implication, all of these objects are red Notice that η 0 in the theorem is green The objective here is to use the green objects to produce the red objects We shall do that next We put: (I δ 0,g = η 0 By the assumption of the theorem g(x is defined for each x in the set ( a η 0, a ( a, a + η 0 (II For each real number ǫ > 0, put δ g (ǫ = min { δ f (ǫ, δ h (ǫ, η 0 } This is a beautiful expression since the red object is expressed in terms of the green objects It remains to prove the red implication (63 using the green implications the assumptions of the theorem To prove (63, assume that 0 < x a < δ g (ǫ Then, clearly, 0 < x a < η 0 This is telling me that x a that x is no further than η 0 from a Consequently, x ( a η 0, a ( a, a + η0 Therefore, by the assumption of the theorem f(x g(x h(x (64

8 6 Subtracting L from each term in this inequality, I conclude that f(x L g(x L h(x L (65 Using the property of the absolute value that u u u for each real number u, we conclude that f(x L f(x L g(x L h(x L h(x L (66 From the assumption 0 < x a < δ g (ǫ, we conclude that 0 < x a < δ f (ǫ By the green implication (6, this implies that f(x L < ǫ therefore ǫ < f(x L (67 From the assumption 0 < x a < δ g (ǫ, we conclude that 0 < x a < δ h (ǫ By the green implication (6, this implies that h(x L < ǫ (68 Putting together the inequalities (66, (67 (68, we conclude that The inequalities in (69 are equivalent to ǫ < g(x L < ǫ (69 g(x L < ǫ (60 This proves that 0 < x a < δ g (ǫ implies g(x L < ǫ this is exactly the red implication (63 This completes the proof The following theorem is the Scissors Squeeze Theorem Theorem 6 Let f, g h be given functions let a R L R Assume that f(x = L, h(x = L, 3 There exists η 0 > 0 such that f(x, g(x h(x are defined for all x ( a η 0, a ( a, a+ η 0 f(x g(x h(x for all x ( a η 0, a, Then h(x g(x f(x for all x ( a, a + η 0 g(x = L

9 7 6 Examples for Squeeze Theorems The following picture the numbers that you can see on it are essential for getting squeezes for its involving trigonometric functions The table to the left shows the numbers that you should be able to identify on the picture Geometric Associated Object Number Circular Arc D from C to B u B Line Segment 0A cosu Line Segment AB sin u Line Segment AC Line Segment CB cos u You Calculate 0 A C Line Segment CD tanu Line Segment 0B Line Segment 0C Example 6 Prove that x 0 cosx = Solution Set η 0 = π Consider positive u Look at the picture above The triangle ACB is a 3 right triangle Therefore its hypothenuse, the line segment CB, is longer than its side AC which equals to cosu Thus cosu = AC CB (6 The line segment CB is a segment of a straight line, therefore it is shorter than any other curve joining C B In particular it is shorter than the circular arc joining the points C B The length of this circular arc is u Thus CB Length of the Circular Arc from C to B ( = u (6 Putting together the inequalities (6 (6, we conclude that cos u u for all 0 < u < π 3 (63 Since the length 0A = cosu is smaller than, from (63 we conclude that 0 cosu u for all 0 < u < π 3,

10 8 or, equivalently, u cosu for all 0 < u < π 3, Now we substitute u = x use the fact that cos x = cos x (6 becomes x cosx for all π 3 < x < π 3 (64 ( This is a swich squeeze for cosx It is easy to prove that = x = x 0 x 0 (Please prove this using the definition! Now the Swich Squeeze Theorem implies that cosx = x 0 Example 6 Prove that x 0 sin x x = Solution To get a swich squeeze for this problem consider the following three areas on the picture above Area The triangle 0CB Area The segment of the unit disc bounded by the line segments 0C 0B the circular arc segment joining points C B Area 3 The triangle 0CD The picture tells clearly the inequality between these areas Write that inequality Calculate each area in terms of the numbers that appear in the table above This will lead to the inequality, which when simplified gives cosu sin u u for all 0 < x < π 3 (65 Using the same idea as in the previous example, the inequality (65 leads to cosx sin x ( x for all x π ( 3, 0 0, π (66 3 The inequality (66 is exactly what we need in the Swich Squeeze Theorem Please fill in all the details of the rest of the proof Example 63 Prove that x 0 cosx x = Solution To establish squeeze inequlaities consider three lengths: Length The line segment AB Length The line segment CB Length 3 The length of a circular arc joining the points C B

11 9 The picture tells clearly the inequalities between these three lengths Write these inequalities Calculate each length in terms of the numbers that appear in the table above This will lead to the inequalities, which, when simplified, give ( sin u u cosu u for all 0 < u < π 3 (67 From the inequality (67 one inequality established in a previous example you can get an easy swich squeeze Please fill in all the details of the rest of the proof Example 64 Prove that x 0 ln( + x x = Solution The idea is to use the definition of ln as an integral work with areas to get squeeze inequalities 63 Algebra of its A nickname that I gave to a function which has a it L when x approaches a is: f is constantish L near a If we are dealing with constant functions f(x = L g(x = K, then clearly the sum f + g of these two functions is a constant function equal to L + K The same is true for the product fg which is the constant function equal to LK Another question is whether we can talk about the reciprocal /f If L 0, then the reciprocal of f is defined it equals /L In this section we shall prove that all these properties hold for constantish functions Theorem 63 Let f, g, h, be functions with domain range in R Let a, K L be real numbers Assume that ( f(x = K, ( g(x = L Then the following statements hold (A If h = f + g, then h(x = K + L (B If h = fg, then h(x = KL (C If L 0 h = f, then g h(x = K L Exercise 63 Use the algebra of its to give much simpler proofs for most of the its in the previous exercises examples

University Calculus I. Worksheet # 8 Mar b. sin tan e. sin 2 sin 1 5. b. tan. c. sec sin 1 ( x )) cos 1 ( x )) f. csc. c.

University Calculus I. Worksheet # 8 Mar b. sin tan e. sin 2 sin 1 5. b. tan. c. sec sin 1 ( x )) cos 1 ( x )) f. csc. c. MATH 6 WINTER 06 University Calculus I Worksheet # 8 Mar. 06-0 The topic covered by this worksheet is: Derivative of Inverse Functions and the Inverse Trigonometric functions. SamplesolutionstoallproblemswillbeavailableonDL,