Limits
0 likes1,161 views
This document discusses limits of functions. It begins by defining the limit of a function f(x) as x approaches a number c as the value that f(x) approaches as x gets closer to c. It provides examples of limits, including one-sided limits and limits at infinity. Key theorems are presented for computing limits, including properties of limits and the sandwich theorem. The document focuses on conceptual understanding and applying techniques to evaluate a variety of limit examples.
Limits
- 1. LIMITS OF FUNCTIONS 1 Introduction The limits concept is one of the most important ideas in calculus. It is the foundation of almost of all mathematical analysis. The limit of a function f (x) as x approaches c is the value that f (x) approaches as x approaches the number c. Definition 1 If f (x) get closer and closer to a number L as x gets closer and closer to c from both sides, then L is the limit of f (x) as x approaches c. Mathematically this is written as lim f (x) = L x→c In limits we look at the behaviour of f (x) as x gets closer and closer to c. We are only talking about what happens to f (x) as x approaches c, and not about what happens when x equals c. It is therefore important to remember that limits describe the behavior of a function near to a particular point, not necessarily at the point. For example: Example 1 : Consider the following function x2 + 2 when x = 2; f (x) = 9 when x = 2 The limit of f (x) as x approaches 2 is equal to 6. Yet the function behave quite differently at x = 2 as f (2) = 9. Suppose we want to evaluate the limit f (x + 5) − f (5) lim , x = 0, x→0 x where f (x) = x2 + 3. A straight substitution of x = 0 leads to the meaning expression f (5) − f (5) 0 = 0 0 1
- 2. and we must penetrate a little deeper. We have f (5) = 28 and f (x + 5) = (x + 5)2 + 3 = x2 + 10x + 28. Therefore, f (x + 5) − f (5) x2 + 10x + 28 − 28 = = x + 10 x x Therefore, f (x + 5) − f (5) lim = lim (x + 10) = 10 x→0 x x→0 Example 2 : Find the limit of f (−1 + h) − f (−1) lim √ , h > 0, h→0 h where f (x) = x3 . Solution We have f (−1) = −1 and f (−1 + h) = (−1 + h)3 = −1 + 3h − 3h2 + h3 . So, f (−1 + h) − f (−1) = h(3 − 3h + h2 ). Hence, f (−1 − +h) − f (−1) h(3 − 3h + h2 ) √ lim √ = lim √ = lim h(3 − 3h + h2 ) = 0 h→0 h h→0 h h→0 2 Limit Applications: Derivatives Definition 2 If f is a function, the derivative of the function f , denoted by f is defined by the fomular f (x + h) − f (x) f (x) = lim h→0 h In this definition, x remain fixed, while h tends to zero. If the limit does not exist for a particular value of x, then the functin has no derivative for that value. We also say that a function f is diffentiable at x if the above limit exists. A function is f is simply said to be differentiable if it has a derivative for all x in its domain. Example 3 : Given f (x) = x2 . Find the f (x). Solution: We have f (x) = x2 and f (x+h) = x2 +2xh+h2 . Therefore, f (x+h)−f (x) = 2xh+h2 . Thus, (x + h) − f (x) 2xh + h2 f (x) = lim = lim = lim (2x + h) = 2x. h→0 h h→0 h h→0 2
- 3. 1 Example 4 : Given that f (x) = x , x = 0. Find f (x). 1 1 x−(x+h) −h Solution: We have f (x − h) − f (x) = (x+h) − x = x(x+h) = x(x+h) . Thus, (x + h) − f (x) −h −1 −1 f (x) = lim = lim = lim 2 = 2. h→0 h h→0 hx(x + h) h→0 (x + xh) x 3 Computation of Limits To compute limits of complex functions we need the following theorems Theorem 3 1. limx→c k = k, where k is any constant 2. limx→c x = c Theorem 4 Suppose that lim f (x) = L and lim g(x) = M. x→c x→c Then 1. limx→c (f (x) + g(x)) = L + M 2. limx→c (f (x) − g(x)) = L − M 3. limx→c kf (x) = kL, where k is a any constant 4. limx→c f (x)g(x) = LM f (x) L 5. limx→c g(x) = M , provided M = 0 6. limx→c (f (x))n = Ln , where n is any positive integer √ 7. limx→c n f (x) = n L Example 5 : Find lim (2x2 − 5x + 3) x→8 3
- 4. Solution lim (2x2 − 5x + 3) = lim (2x2 − 5x) + lim 3 x→8 x→8 x→8 2 = lim (2x ) − lim (5x) + lim 3 x→8 x→8 x→8 = 2 lim (x2 ) − 5 lim x + lim 3 x→8 x→8 x→8 2 = 2 lim x − 5 lim + lim 3 x→8 x→8 x→8 2 = 2(8) − 5(8) + 3 = 91 Example 6 : Find x3 + 5 lim x→−2 x2 + x − 3 Solution The limit of the denominator is limx→−2 (x2 + x − 3) = (−2)2 − 2 − 3 = −1, which is not zero. So, we can use rule 5 in Theorem 4. Thus, x3 + 5 lim x → −2(x3 + 5) lim = x→−2 x2 + x − 3 lim x → −2(x2 + x − 3) (−2)3 + 5 = −1 −3 = −1 = 3. Example 7 : Find x2 − 8x + 15 lim x→3 x2 + 4x − 21 Solution: The limit of the denominator is lim (x2 + 4x − 21) = (3)2 + 4(3) − 21 = 0. x→−3 So we cannot use rule 5 in Theorem 4. Instead, we if it is possible to simplify the given function. A standard method is to factorize both numerator and denomenator. We have x2 − 8x + 15 = 4
- 5. (x − 3)(x − 5) and x2 + 4x − 21 = (x − 3)(x + 7). Thus, x2 − 8x + 15 (x − 3)(x − 5) lim = lim x→3 x2 + 4x − 21 x→3 (x − 3)(x − 5) (x − 5) = lim (Cancelling the common factor x − 3) x→3 (x − 5) 3−5 = 3+7 1 = − 5 Example 8 : Find √ 3x + 1 − 4 lim x→5 x−5 0 Solution The limit of the function is of the form 0 . The main problem here is it might not very clear how th find the hiden factor x − 5 of in the numerator. One techniques is to rationalize. That is, we use the algebraic identity (a + b)(a − b) = a2 − b2 Thus, √ √ √ 3x + 1 − 4 ( 3x + 1 − 4)( 3x + 1 + 4) lim = lim √ x→5 x−5 x→5 (x − 5)( 3x + 1 + 4) √ ( 3x + 1)2 − 16 = lim √ x→5 (x − 5)( 3x + 1 + 4) 3x + 1 − 16 = lim √ x→5 (x − 5)( 3x + 1 + 4) 3x − 15 = lim √ x→5 (x − 5)( 3x + 1 + 4) 3(x − 5) = lim √ x→5 (x − 5)( 3x + 1 + 4) 3 = lim √ x→5 3x + 1 + 1 3(x − 5) = lim x→5 (x − 5)( 3(5) + 1 + 1 3 = lim x→5 3(5) + 1 + 1 3 = 8 5
- 6. Example 9 : Find 1 lim x2 sin x→0 x Solution: Although the function is of the form f (x)g(x), we cannot use rule 5 in Theorem 4, 1 1 because limx→0 sin x doesnot exist. To see this, as x getting smaller and smaller x keep growing in size indefinetly!! In this case we need the following theorem to accomplish the task. Theorem 5 (Sandwitch Theorem) Suppose that f (x), g(x) and h(x) have property that f (x) ≤ g(x) ≤ (h) for all x close to c, except possibly for x = c. If limx→c f (h) = L and limx→c h(x) = L then limx→c g(x) = L We know 1 −1 ≤ sin ≤1 x Thus, 1 −x2 ≤ x2 sin≤ x2 x Now limx→0 −x2 = 0 and limx→0 x2 = 0. Therefore, by the Sandwitch theorem, we have 1 lim x2 sin =0 x→0 x 4 One Sided Limits Sometimes we want to study where f (x) is heading when x is heading toward a only from the right hand side of a or only from the left hand side of c; and so we have the following notions of one-sided limits: We write lim f (x) = L x→a+ to mean that as x is moving toward a from the right hand side of c; the function value f (x) approaches L. We also say that the right hand limit of f as x approaches c from the right is L, and L is the right hand limit of f (x) at c. We write lim f (x) = L x→a− 6
- 7. to mean that as x is moving toward a from the left hand side of a, the function value f (x) ap- proaches L. We also say that the right hand limit of f as x approaches c from the left is L, and L is called the left hand limit of f (x) at c. It is not difficult to see that rules of limit calculations can be also applied to one-sided limits. Example 10 : For the function 1 − x2 if x < 2 f (x) = 2x + 1 if x ≥ 2 Find lim f (x) and lim f (x) x→2− x→2+ Solution Since f (x) = 1 − x2 for x < 2, we have lim f (x) = lim (1 − x2 ) = −3 x→2− − x→2 Similarly, f (x) = 2x + 1 if x ≥ 2, so lim f (x) = lim (2x + 1) = 5 x→2+ + x→2 Example 11 : For the function x(x − 3) if x < 4 f (x) = x2 x+4 if x ≥ 4 Find lim f (x) and lim f (x) x→4− x→4+ Solution Since f (x) = x(x − 3) for x < 4, we have lim f (x) = lim (x(x − 3)) = 4 x→4− − x→2 x2 Similarly, f (x) = x+4 if x ≥ 4, so x2 42 lim f (x) = lim = =2 x→4 + + x→4 x+4 8 7
- 8. Theorem 6 If f is a function and c and L are numbers, then lim f (x) = L x→c if and only if lim f (x) = L and lim f (x) = L. x→c− x→c+ The theorem says that the limit of f exists of and only if both sided limits exist and they have the same value. Example 12 : Determine whether limx→1 f (x) exists, where x+1 ifx < 1 f (x) = −x2 + 4x − 1 if x ≥ 1 Solution Computing the one-sided limits at x = 1, we find lim f (x) = lim (x + 1) = (1) + 1 = 2 x→1− − x→1 and lim f (x) = lim (−x2 + 4x + 1) = −(1)2 + 4(1) + 1 = 2 + + x→1 x→1 Since the two one-sided limits are equal, it follows that the limit of f (x) at 1 exist and we have lim f (x) = lim f (x) = lim f (x) = 2. x→1− + x→1 x→1 5 Limits at Infinity The symbols ∞ and −∞ are not regarded as real numbers. They are symbols to indicate that a number increases or decreases indefinitely. When the limit of a function is ∞ or −∞ no limit exists; the symbol is used for convenience only. Theorem 7 1 1 1 1 lim = 0, lim = 0, lim = 0, and lim = 0 for any positive integer p x→∞ x x→+∞ x x→−∞ x x→∞ xn Example 13 : Evaluate 3x − 2 lim x→∞ 5x + 4 8
- 9. Solution: 3x−2 3x − 2 x lim = lim x→∞ 5x + 4 x→∞ 5x+4 x 3x −2 = lim 5x x x 4 x→∞ x +x 2 3− x = lim x→∞ 5 + 4 x 3−0 = 5+0 3 = 5 Example 14 : Find √ x2 − 1 lim x→∞ 2x + 1 Solution: √ 1− 1 x2 − 1 x2 lim = lim 1 x→∞ 2x + 1 2+x→∞ x √ 1−0 = 2+0 1 = 2 Example 15 : Find 3x + 2 lim x→−∞ x2−x+1 Solution : 3x 2 3x + 2 x2 + x2 lim 2−x+1 = lim x2 x 1 x→−∞ x x→−∞ − + x2 x2 x2 3 2 x + x2 = lim 1 1 1− x→−∞ x + x2 0+0 = 1−0+0 = 0 9