Sections 2.1, 2.2 and 2.4: Limit of a function Motivation:

Transcription

1 Sections 2.1, 2.2 and 2.4: Limit of a function Motivation: There are expressions which can be computed only using Algebra, meaning only using the operations +,, and. Examples which can be computed using algebra: a) Area of a rectangle b a b) Area of a circle A = a b. r c) Area of a triangle A = πr 2. h b A = 1 2 hb. 1

2 2 d) Average speed: A stone which is dropped from a 200 m tall tower has the height h(t) = t 2 [m] after t seconds. What is its average velocity between the time t 1 = 1s and t 2 = 3s? h(3) h(1) v ave = 3 1 = ( ) ( ) = 2 e) Consider the function f(x) = 1 2 x2 = 19.6 [ m s What is the slope of the secant through the points on the graph A(1,.5) and B(3, 4.5)? slope = y f(3) f(1) = = x = 2. y B ]. secant y A x x

3 Objects which cannot be computed using Algebra. a) Find the area between the x-axis, the graph of a function y = f(x), x = a, and x = b, 3 a b b) Instantaneous velocity. A stone which is dropped from a 200 m tall tower has the height h(t) = t 2 [m] after t seconds. What is its instantaneous velocity a time t 1 = 1s?

4 4 Purpose of Calculus: To compute expressions which cannot be computed using Algebra, but need to be first approximated by algebraic expressions and then be computed, by Taking Limits. The purpose of this class (and the next one) will be to understand what taking limits exactly means.

5 We start with two examples. Example. The function f(x) = x2 9 x 3, is defined for all values x 3 and can be simply computed there by inserting the value of x. Question: We are interested in what happens if we insert numbers which are getting closer and closer to the value 3? Inserting some numbers close to 3, we get the following table. x = f(x) = Guess: f(x) gets closer and closer to 6 if x gets closer and closer to 3. Proof. For any number x 3 we can cancel: 5 f(x) = x2 9 (x 3)(x + 3) = = x + 3. x 3 x 3 Thus f(x) = x + 3, for all numbers x 3. So we can conclude: f(x) approaches the value 6 when x approaches 3. We will write lim f(x) = 6. x 3 Note: the functions f(x) = x2 9 x 3 and g(x) = x + 3 are not equal functions, but they have equal function values for all numbers x 3.

6 6 Example. Consider the function f(x) = 1 2 x2 What is the slope of the tangent through the point (1,.5)? y secant y tangent x x We compute the slope of the secant going through the points (1, f(1)) = (1,.5) and some other point (1 + h, f(1 + h)) = (1 + h, 1 2 (1 + h)2 ), with h 0: x = (1 + h) 1 = h y = 1 2 (1 + h)2 1 2 = h h2 = h h2. Slope of secant y x = h h2 = h 2 h. Now we let h approach 0, and thus the point (1+h, f(1+h)) approach (1,.5), then the slope of the secant approaches the value 1. Thus we conclude that the slope of the tangent to the function f(x) = 1 2 x2 has the value 1.

7 7 Theory: In Mathematics, as in any area of Sciences, we need to be precise, and only use words which have an exact meaning: For example one variable approaching a number L does not have an exact meaning. Definition 1: The Limit of a Function Let f(x) be a function defined on an open interval containing a number a with exception of possibly at a itself. We say that f(x) approaches the value L if x approaches a and we write: lim f(x) = L, x a if the following is true: for all numbers ε > 0 there is a number δ > 0, so that the following is true: If 0 < x a < δ then f(x) L < ε. (Problems 1 and 4) y f(x) L + ε L L ε a δ a a + δ x Explanation of picture: lim x a f(x) = L. For all x which lie in the interval (a δ, a + δ), its image f(x) lies in the interval L ε, L + ε. We say this δ works for for ε.

8 8 Also in the following picture: lim x a f(x) = L (even if f(a) is not defined) y f(x) L a x In the following picture: lim x a f(x) does not exist y f(x) L a x

9 9 Right hand and Left hand limits: Definition 2: Left hand limi Let f(x) be a function defined on an open interval (c, a), with c < a. We say that f(x) approaches the value L if x approaches a from the left and we write: lim f(x) = L, x a if for all number ε > 0 there is a number δ > 0 so that the following is true: If a δ < x < a then f(x) f(a) < ε. Definition 3: Right hand limit Let f(x) be a function defined on an open interval (a, c), with c > a. We say that f(x) approaches the value L if x approaches a from the right and we write: lim f(x) = L, x a + if for all number ε > 0 there is a number δ > 0 so that the following is true: If a < x < a + δ then f(x) f(a) < ε. lim x a f(x) = L and lim x a + f(x) = M. y f(x) M L a x

10 10 Theorem 1: Assume that the function y = f(x) is defined on an open interval which contains a point a, with exception of possibly at a. Then, for a number L we have: if and only if lim f(x) = L x a lim f(x) = L and lim x a + f(x) = L. x a Theorem 2: Assume that the function y = f(x) is defined on an open interval which contains a point a, with exception of possibly at a. Then, for a number L we have: if and only if lim f(x) = L x a lim f(x) L = 0. x a y f(x) L f(x) L a x

11 11 Two Examples for which lim x a f(x) does not exist y x For f(x) = 1 x, lim x 0 f(x) does not exists. We can be more precise 1 lim = + and x 0 + x lim x 0 1 x =.

12 12 Second example: f(x) = sin(1/x). Note that if x approaches 0 sin(1/x) runs through all numbers between 1 and 1 infintily many times, and thus does approach any single number. Thus lim sin(1/x) does not exist. x 0 y... x

13 13 Problems: Problem 1. Using the definition of limits show that lim 3x 1 = 5. x 2 Solution: Let ε > 0 be arbitrary. We need to find a number δ > 0 (depending on ε) which works, meaning that the following implication must be true: If 0 < x 2 < δ then (3x 1) 5 < ε. Or in short 0 < x 2 < δ (3x 1) 5 < ε. Note that (3x 1) 5 = (3x 6 = 3 x 2 < ε (the idea is to write the wanted difference (f(x) L in terms of x a ) Now ask yourself: how small has x 2 to be in order for 3 x 2 to be smaller than ε? Well, easy, it has to be smaller than ε 3. Thus δ = ε 3 should work. Indeed, let us choose δ = ε 3, and check that this δ works: If 0 < x 2 < δ, then If 0 < x 2 < δ then (3x 1) 5 = 3x 6 = 3 x 2 < 3δ = ε. Remark. You might have the impression that we did the the same thing twice, and that it was superfluous to do the last step to check that our choice of δ works. Given the simplicity of the example, it was indeed clear. But for more complicated examples this is not the case. Actually, in order to verify that a certain δ > 0 works we do not show how we got it. We need to define it, and then verify that it works, meaning that the implication holds. If 0 < x a < δ then f(x) f(a) < ε

14 14 Problem 2. Using the definition of limits show that lim 5 2x = 1. x 3

15 15 Let us now consider nonlinear functions: Problem 3. Let f(x) = x 3 2, and let ε =.1. Find δ > 0 (not necessarily optimal) which works, meaning, it should satisfy the following implication: 0 < x 2 < δ then f(x) 6 <.1. Solution: We want to find δ > 0 so that We compute: 0 < x 2 < δ x <.1. x = x 3 8 = (x 2) (x 2 + 2x + 4) = x 2 x 2 + 2x + 4. To find a δ we can argue like that: as long as we make sure that δ 1, then for any x with x 2 < δ it follows that x 3 and thus x 2 + 2x + 4 < = 19. Thus, if we choose δ =.1/19 = 1/190, then that choice will work. Indeed, if 0 < x 2 < then x = x 3 8 = (x 2) (x 2 +2x+4) = x 2 x 2 +2x+4 < =.1.

16 16 Problem 4. Using the definition of limits show that lim x 2 3x2 5 = 7. Solution: Let ε > 0, we need to find δ. Note that (as before we try to factor out x a, with a = 2): (3x 2 5) 7 = 3x 2 12 = 3 x 2 4 = x 2 3 x + 2. We argue as before: as long as we make sure that δ 1, it follows that x 2 < δ implies that x < 3 and, thus, for the above factor of x 2 : 3 x + 2 < 15. So we should choose δ = ε δ = min(1, ε 15 ). 15 or δ = 1, which ever is smaller, in other words Verification that this choice works: if x 2 < δ, then: (3x 2 5) 7 = 3x 2 12 = 3 x 2 4 = x 2 3 x + 2 < ε 15 = ε. 15

17 17 Problem 5. State the precise definition of what it means that lim f(x) =, x a and using that definition show that lim x 0 1 x 2 =. Solution: Assume that the function y = f(x) is defined on an open interval containing a, but not at a itself. We write lim x a f(x) =, if for any number C > 0 there exists a δ > 0 so that If 0 < x a < δ, then f(x) > C. Proof that lim x 0 1 x 2 =. Let C > 0. For ensuring that 1 x 2 > C, it is enough to assume that x < 1 C. Thus, δ = 1 C will work. Indeed, If x 0 = x < δ = 1 C, then ) 2 ( ) 2 = = C. ( 1 x 2 > 1 δ 1 1 C

18 18 Problem 6. Only using the definition of limits prove Theorem 1.

19 19 Problem 7. For a > 0, show that x = a. lim x a Solution: Let a > 0 and ε > 0 be given. We need to find δ > 0 (possibly depending on a and ε) so that if 0 < x a < δ then x a < ε. We compute for x > 0: x a ( x a) ( x + a) = x + a = x a x a. x + a a So, first δ has to be chosen small enough so that x a < δ implies that x 0 (and thus x is in the range of ) so δ should be at most a. Secondly, from above inequalities δ should be smaller than ε a. Thus let us choose δ = min ( a, ε a ), and verify that this choice works: if a x < δ then x a ( x a) ( x + a) = x + a = x a x + a x a a < δ a ε.