Chapter 1: Limits and Continuity

Chapter 1: Limits and Continuity Winter 2015 Department of Mathematics Hong Kong Baptist University 1/69

1.1 Examples where limits arise Calculus has two basic procedures: differentiation and integration. Both procedures are based on the fundamental concept of the limit of a function. It is the idea of limit that distinguishes Calculus from Algebra, Geometry, and Trigonometry, which are useful for describing static situations. This section considers some examples of phenomena where limits arise in a natural way. 2/69

Example 1: Consider the area of a circle of radius r. In high school we are taught that the area A of the circle is A = πr 2. The deduction of this area formula lies in regarding the circle as a limit of regular polygons. Suppose a regular polygon having n sides is inscribed in the circle of radius r, and let A n be the area of the polygon. It is clear that A n is less than A for each n. But if n is large, A n is close to A. That is, we would expect that A n approaches the limit A when n goes to infinitely large. 3/69

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The area A n of the polygon has the following expression: A n = r 2 n sin( π n )cos(π n ). Then by the trigonometric formula, A n = r2 n 2 sin(2π n ). Finally, noting that sin(t)/t 1 as t 0, we have A n = πr 2 1 2π/n sin(2π/n) πr2 as n. 5/69

1.2 Limits of Functions The concept of limit is the cornerstone on which the development of Calculus rests. Intuitively speaking, the limit process involves examining the behavior of a function f(x) as x approaches a number c that may or may not be in the domain of f. 6/69

Example 2 Describe the behavior of the function f(x) = x2 4 x 2 near x = 2. Solution: Note that f(x) is defined for all real numbers x except x = 2. For any x 2 we have f(x) = (x +2)(x 2) x 2 = x +2 for x 2. The graph of f is the line y = x +2 with the point (x,y) = (2,4) removed. Though f(2) is not defined, it is clear that f(x) becomes closer to 4 as x approaches 2. In other words, we say that f(x) approaches the limit 4 as x approaches 2. We write this as x 2 4 lim f(x) = lim x 2 x 2 x 2 = 4. 7/69

Graph of f(x) 8/69

(Informal) definition of the limit Definition If f(x) is defined for all x near c, except possibly at c itself, and if we can ensure that f(x) is as close as we want to L by taking x close enough to (but different from) c, we say that the function f approaches the limit L as x approaches c, and we write it as lim f(x) = L. x c 9/69

Limits of two linear functions (1) For the identity function f(x) = x, we have lim f(x) = lim x = c. x c x c That is, the limit of f(x) = x is c as x approaches c. 10/69

Limits of two linear functions (1) For the identity function f(x) = x, we have lim f(x) = lim x = c. x c x c That is, the limit of f(x) = x is c as x approaches c. (2) For any constant k, we have lim k = k. x c That is, the limit of a constant is the constant itself. 11/69

Definition (a) If f(x) is defined on some interval (a,c) extending to the left of x = c, and if we can ensure that f(x) is as close as we want to L by taking x to the left of c and close enough to c, then we say f(x) has left limit L at x = c, and we write it as lim = L. x c f(x) (b) If f(x) is defined on some interval (c,b) extending to the right of x = c, and if we can ensure that f(x) is as close as we want to L by taking x to the right of c and close enough to c, then we say f(x) has right limit L at x = c, and we write it as lim = L. x c +f(x) 12/69

Example 3 (a) Find the left and right limits of f(x) = x2 4 at x = 2; x 2 (b) Find the left and right limits of g(x) = x/ x at x = 0. Solution: 13/69

Example 3 (a) Find the left and right limits of f(x) = x2 4 at x = 2; x 2 (b) Find the left and right limits of g(x) = x/ x at x = 0. Solution: (a) For f(x), we have lim = 4 and x 2 f(x) lim x 2 +f(x) = 4. 14/69

Example 3 (a) Find the left and right limits of f(x) = x2 4 at x = 2; x 2 (b) Find the left and right limits of g(x) = x/ x at x = 0. Solution: (a) For f(x), we have lim = 4 and x 2 f(x) lim x 2 +f(x) = 4. (b) For g(x), we have lim = 1 and x 0 g(x) lim = 1. x 0 +g(x) Note that this is the so-called sign function, i.e., sgn(x) = x/ x. 15/69

Relationship between one-sided and two-sided limits Theorem [Existence of a Limit] The two-sided limit limf(x) exists if and x c only if both one-sided limits lim and lim x c f(x) x c +f(x) exist and are equal to each other. In this case, we have lim x c f(x) = lim = lim x c f(x) x c +f(x). Note: If the two one-sided limits do not agree, or if one of them does not exist, then the two-sided limit does not exist. For example, lim x 0 x x does not exist. 16/69

Limit Rules If lim x c f(x) = L, lim x c g(x) = M, and k is a constant, then 1) lim x c kf(x) = kl. 2) lim x c [f(x)±g(x)] = L±M. 3) lim x c [f(x) g(x)] = LM. f(x) 4) lim x c g(x) = L M if M 0. (5) If f(x) g(x) on an interval containing c in its interior, then the order of the limits is preserved, i.e., L M. 17/69

Remark 1: For any positive integer n, by property 3) we have lim x c [f(x)]n = L n. More generally, for any non-integer t such that L t exists, we still have lim x c [f(x)]t = L t. Remark 2: Let f(x) = x, we have lim x c x n = c n. Let P(x) and Q(x) be any two polynomials, then (i) lim x c P(x) = P(c), (ii) lim x c P(x) Q(x) = P(c) Q(c) if Q(c) 0. 18/69

Example 4: Find lim x 0 x(1+x 1 ). 19/69

Example 4: Find lim x 0 x(1+x 1 ). Wrong answer: lim x 0 x(1+x 1 ) = lim x lim(1+x 1 ) = 0 =? x 0 x 0 Correct answer: lim x 0 x(1+x 1 ) = lim(x +1) = lim x +1 = 1. x 0 x 0 20/69

x 2+x Example 5: Find lim. x 2 x 2 21/69

x 2+x Example 5: Find lim. x 2 x 2 Wrong answer: x 2+x lim x 2 x 2 = lim x 2(x 2+x) lim x 2 (x 2) = 0 0 =? 22/69

x 2+x Example 5: Find lim. x 2 x 2 Correct answer: x 2+x lim x 2 x 2 (x 2+x)(x + 2+x) = lim x 2 (x 2)(x + 2+x) x 2 x 2 = lim x 2 (x 2)(x + 2+x) (x 2)(x +1) = lim x 2 (x 2)(x + 2+x) x +1 = lim x 2 x + 2+x = 3 4. 23/69

The Squeeze Theorem Theorem Suppose that f(x) g(x) h(x) holds for all x in some open interval containing a, except possibly at x = a itself. Suppose also that lim f(x) = lim h(x) = L. x a x a Then lim x a g(x) = L too. Similar statements hold for left and right limits. 24/69

The Squeeze Theorem 25/69

Example 6 (a) Given that 3 x 2 u(x) 3+x 2 for all x 0, find lim u(x). x 0 (b) If lim f(x) = 0, find lim f(x). x a x a Solution: 26/69

Example 6 (a) Given that 3 x 2 u(x) 3+x 2 for all x 0, find lim u(x). x 0 (b) If lim f(x) = 0, find lim f(x). x a x a Solution: (a) Note that lim(3 x 2 ) = 3 and lim(3+x 2 ) = 3. By the x 0 x 0 Squeeze theorem, we have lim u(x) = 3. x 0 27/69

Example 6 (a) Given that 3 x 2 u(x) 3+x 2 for all x 0, find lim u(x). x 0 (b) If lim f(x) = 0, find lim f(x). x a x a Solution: (a) Note that lim(3 x 2 ) = 3 and lim(3+x 2 ) = 3. By the x 0 x 0 Squeeze theorem, we have lim u(x) = 3. x 0 (b) Note that f(x) f(x) f(x) and lim{ f(x) } = lim f(x) = 0. x a x a By the Squeeze theorem, we have lim x a f(x) = 0. 28/69

Example 7 ( ) 1 Calculate lim x sin. x 0 x Solution: 29/69

Example 7 ( ) 1 Calculate lim x sin. x 0 x Solution: Note that sin(1/x) 1 for all values of x. Thus, we have ( ) 1 x x sin x. x Since lim x 0 x = lim x 0 x = 0, by the Squeeze theorem, we get ( ) 1 lim x sin = 0. x 0 x 30/69

( ) 1 f(x) = x sin x x*sin(1/x) 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x 31/69

( ) 1 Note: lim sin does NOT exist! x 0 x 2 1.5 1 0.5 sin(1/x) 0 0.5 1 1.5 2 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x 32/69

Example 8 (See also 2.5) Calculate lim θ 0 sinθ θ Solution: (where θ is in radians). 33/69

Example 8 (See also 2.5) Calculate lim θ 0 sinθ θ Solution: For 0 < θ < π/2, (where θ is in radians). Area of sector = θ 2 Area of OPB = 1 2 sinθcosθ Area of OTA = 1 2 tanθ Hence, sinθcosθ 2 θ 2 tanθ 2. 34/69

Example 8 (See also 2.5) Calculate lim θ 0 sinθ θ Solution: (where θ is in radians). (i) Thus, for 0 < θ < π/2, we have cosθ sinθ θ 1 cosθ with lim θ 0 +cosθ = 1, lim 1 θ 0 + cosθ = 1. By the Squeeze theorem, we obtain lim θ 0 + sinθ θ = 1. 35/69

Example 8 (See also 2.5) Calculate lim θ 0 sinθ θ Solution: (where θ is in radians). (ii) For θ < 0, we can substitute α = θ to get sinθ sin( α) sin(α) lim = lim = lim = 1. θ 0 θ α 0 + ( α) α 0 + α (iii) Thus, combining the two one-sided limits, we get sinθ lim θ 0 θ = 1. Note: This is an important result and should be memorized. 36/69

Example 8 Calculate lim x 0 1 cos(x) x 2. Solution: 37/69

Example 8 Calculate lim x 0 1 cos(x) x 2. Solution: Using the trigonometric identity cos(x) = 1 2sin 2 (x/2), we deduce that 1 cos(x) 2sin 2 (x/2) lim x 0 x 2 = lim x 0 x 2 2sin 2 (x/2) = lim x 0 = 1 2 ( lim x 0 4 (x/2) 2 sin(x/2) (x/2) ) ( lim x 0 ) sin(x/2) = 1 (x/2) 2. 38/69

1.3 Limits at Infinity and Infinite Limits In this section, we consider several scenarios not covered by the limit definitions in Section 1.2: (i) limit at infinity: when x becomes arbitrarily large positive (denoted by x ). Remark: The symbol, called infinity, does not represent a real number. We cannot use in arithmetic in the usual way. For instance, we cannot say +1 >. (ii) limit at negative infinity: when x becomes arbitrarily large negative (denoted by x ). (iii) infinite limits: they are not really limits at all but provide useful symbolism for describing the behavior of functions whose values become arbitrarily (positive or negative) large. 39/69

Definition (a) If f(x) is defined on an interval (a, ) and if we can ensure that f(x) is as close as we want to the number L by taking x large enough, then we say that f(x) approaches the limit L as x approaches infinity, and we write lim f(x) = L. x (b) If f(x) is defined on an interval (,b) and if we can ensure that f(x) is as close as we want to the number M by taking x negative and large enough in absolute value, then we say that f(x) approaches the limit M as x approaches negative infinity, and we write lim f(x) = M. x 40/69

Example 9 x (a) Evaluate lim f(x) and lim f(x) for f(x) = x x (b) Describe the behavior of g(x) = x2 5 x 2 Solution: x 2 +1. near x = 2. 41/69

Example 9 x (a) Evaluate lim f(x) and lim f(x) for f(x) = x x (b) Describe the behavior of g(x) = x2 5 x 2 Solution: (a) For f(x), we guess that lim f(x) = 1 and lim x x 2 +1. near x = 2. f(x) = 1. x The horizontal lines y = 1 and y = 1 are called horizontal asymptotes of the graph f(x). 42/69

Example 9 x (a) Evaluate lim f(x) and lim f(x) for f(x) = x x (b) Describe the behavior of g(x) = x2 5 x 2 Solution: (a) For f(x), we guess that lim f(x) = 1 and lim x x 2 +1. near x = 2. f(x) = 1. x The horizontal lines y = 1 and y = 1 are called horizontal asymptotes of the graph f(x). (b) For g(x), we guess that lim = and x 2 g(x) lim x 2 +g(x) =. The vertical line x = 2 is called a vertical asymptote of the graph g(x). 43/69

Graphs of f(x) and g(x) f(x) g(x) y 1.0 0.5 0.0 0.5 1.0 4 2 0 2 4 x y 10 0 10 20 4 2 0 2 4 x 44/69

Example 10 x Reevaluate lim f(x) and lim f(x) for f(x) = x x Solution: Note that x 2 +1. We have f(x) = x = x 2 (1+ 1 ) x 2 x x (1+ 1 ) x 2 = sgn(x). 1+ 1 x 2 lim f(x) = lim x sgn(x) x lim x 1+ 1 Similarly, we have lim f(x) = 1. x x 2 = 1 1 = 1. 45/69

Limits at infinity for rational functions Corollary Let P m (x) = a m x m + +a 0 and Q n (x) = b n x n + +b 0 be polynomials of degree m and n, respectively, so that a m 0 and b n 0. Then P m (x) lim x Q n (x) and P m (x) lim x Q n (x) (a) equals zero if m < n, (b) equals a m b n if m = n, (c) does not exist if m > n. Or equivalently, we say that the limit is or. 46/69

Example 11: Find the following limits: (a) lim x Solution: 4x 2 x +3 3x 2, (b) lim +5 x 4x 3 4x 2 3 3x 2, (c) lim +5 x 3x +5. 47/69

Example 11: Find the following limits: (a) lim x Solution: 4x 2 x +3 3x 2, (b) lim +5 x 4x 3 4x 2 3 3x 2, (c) lim +5 x 3x +5. 4x 2 x +3 4 x 1 +3x 2 (a) lim x 3x 2 = lim +5 x 3+5x 2 = 4 3. 48/69

Example 11: Find the following limits: (a) lim x Solution: 4x 2 x +3 3x 2, (b) lim +5 x 4x 3 4x 2 3 3x 2, (c) lim +5 x 3x +5. 4x 2 x +3 4 x 1 +3x 2 (a) lim x 3x 2 = lim +5 x 3+5x 2 = 4 3. 4x 3 (b) lim x 3x 2 +5 = lim 4x 1 3x 2 x 3+5x 2 = 0 3 = 0. 49/69

Example 11: Find the following limits: (a) lim x Solution: 4x 2 x +3 3x 2, (b) lim +5 x 4x 3 4x 2 3 3x 2, (c) lim +5 x 3x +5. 4x 2 x +3 4 x 1 +3x 2 (a) lim x 3x 2 = lim +5 x 3+5x 2 = 4 3. 4x 3 (b) lim x 3x 2 +5 = lim 4x 1 3x 2 x 3+5x 2 = 0 3 = 0. 4x 2 3 (c) lim x 3x +5 = lim 4x 3x 1 x 3+5x 1 = 3 =. 50/69

Example 12: Find the limit lim ( x 2 +x x). x Solution: By rationalizing the expression we have lim ( x 2 ( x +x x) = lim 2 +x x)( x 2 +x +x) x x x 2 +x +x = lim x = lim x = 1 2. x x 2 +x +x 1 1+ 1 x +1 51/69

Example 13: Find the following limits, x +1 (a) lim x 2 x 2 Solution:, (b) lim x 1 x 2 1 x 1 x 2, (c) lim 3x +2 x 1 x 1. 52/69

Example 13: Find the following limits, x +1 (a) lim x 2 x 2 Solution: (a) lim x 2 x +1 x 2, (b) lim x 1 does not exist since x 2 1 x 1 x 2, (c) lim 3x +2 x 1 x 1. x +1 x +1 lim = and lim x 2 x 2 x 2+ x 2 =. 53/69

Example 13: Find the following limits, x +1 (a) lim x 2 x 2 Solution: (a) lim x 2 x +1 x 2, (b) lim x 1 does not exist since x 2 1 x 1 x 2, (c) lim 3x +2 x 1 x 1. x +1 x +1 lim = and lim x 2 x 2 x 2+ x 2 =. x 2 1 (b) lim x 1 x 2 3x +2 = lim x +1 x 1 x 2 = 2 1 = 2. 54/69

Example 13: Find the following limits, x +1 (a) lim x 2 x 2 Solution: (a) lim x 2 x +1 x 2 (b) lim x 1, (b) lim x 1 does not exist since x 2 1 x 1 x 2, (c) lim 3x +2 x 1 x 1. x +1 x +1 lim = and lim x 2 x 2 x 2+ x 2 =. x 2 1 x 2 3x +2 = lim x 1 x 1 (c) lim x 1 x 1 = lim 1 = 1 x 1 x +1 2. x +1 x 2 = 2 1 = 2. 55/69

1.4 Continuity Definition We say that a function f is continuous at an interior point c of its domain if lim f(x) = f(c). x c If either lim x c f(x) fails to exist or it exists but is not equal to f(c), then we will say that f is discontinuous at c. 56/69

Example 14 Discuss the continuity of each of the following functions: Solution: (a) f(x) = 1 x, (b) g(x) = x2 1 x +1, (c) h(x) = x. 57/69

Example 14 Discuss the continuity of each of the following functions: Solution: (a) f(x) = 1 x, (b) g(x) = x2 1 x +1, (c) h(x) = x. (a) f(x) is continuous everywhere except for x = 0. 58/69

Example 14 Discuss the continuity of each of the following functions: Solution: (a) f(x) = 1 x, (b) g(x) = x2 1 x +1, (c) h(x) = x. (a) f(x) is continuous everywhere except for x = 0. (b) g(x) is continuous everywhere except for x = 1. 59/69

Example 14 Discuss the continuity of each of the following functions: Solution: (a) f(x) = 1 x, (b) g(x) = x2 1 x +1, (c) h(x) = x. (a) f(x) is continuous everywhere except for x = 0. (b) g(x) is continuous everywhere except for x = 1. (c) h(x) is continuous at any point in the domain of (0, ). 60/69

Properties of continuous functions Assume that f(x) and g(x) are both continuous at point x = c. Then 1) kf(x) is continuous at c, where k is any number. 2) f(x)±g(x) is continuous at c. 3) f(x)g(x) is continuous at c. 4) f(x)/g(x) is continuous at c, if g(c) 0. 5) f(x) is continuous at c, if f(c) > 0. 6) f g = f(g(x)) is continuous at c, if f(x) is also continuous at x = g(c). 61/69

Continuity of basic elementary functions All polynomials are continuous everywhere. sin(x), cos(x), arctan(x) and e x are continuous everywhere. n x is continuous for all x (, ) when n is odd, and for x > 0 when n is even. ln(x) is continuous on x (0, ). arcsin(x) and arccos(x) are continuous on x ( 1,1). x is continuous everywhere. 62/69

Example 15: Discuss the continuity of Solution: h(x) = tan ( 1 1 x ). h(x) can be written as h(x) = f(g(x)) with f(y) = tan(y), g(x) = 1 1 x g(x) is continuous at all points except at x = 1 f(y) is continuous at all points, except where cos(y) = 0, i.e., whenever y = (2k +1) π 2, where k is an integer If h is discontinuous at x = c, then either g discontinuous at x = c, or f is discontinuous at y = g(c) Thus, h is continuous everywhere, except at x = 1 and at 1 1 x = (2k +1)π 2 x = 1 2 (2k +1)π for any integer k. 63/69

Right and left continuity Definition (a) We say that f is right continuous at c if lim x c +f(x) = f(c). (b) We say that f is left continuous at c if lim = f(c). x c f(x) 64/69

Continuity on an interval Definition (a) A function f(x) is said to be continuous on an open interval (a,b) if it is continuous at each point x = c in that interval. (b) A function f(x) is said to be continuous on the closed interval [a,b] if it is continuous on (a,b) and lim x a = f(a) and lim = f(b). +f(x) f(x) x b Or equivalently,... if it is continuous on (a,b), right continuous at a, and left continuous at b. 65/69

The Max-Min Theorem Theorem If f(x) is continuous on the closed, finite interval [a,b], then there exist numbers p and q in [a,b] such that for all x in [a,b], f(p) f(x) f(q). Thus f has the absolute minimum value m = f(p), taken on at the point p, and the absolute maximum value M = f(q), taken on at the point q. Remark: The theorem merely asserts that the minimum and maximum values exist; it does not tell us how to find them. Some techniques for calculating the minimum and maximum values will be introduced in Chapter 4. 66/69

Intermediate Value Theorem Theorem [Intermediate Value Theorem] Suppose that f(x) is continuous on [a,b] and W is any number between f(a) and f(b). Then, there is at least one number c [a,b] for which f(c) = W. Remark: In other words, a continuous function attains all values between any two of its values. For instance, a girl who weighs 5 pounds at birth and 100 pounds at age 12 must have weighed exactly 50 pounds at some time during her 12 years of life, since her weight is a continuous function of time. 67/69

Intermediate Value Theorem Corollary Suppose that f(x) is continuous on [a,b] and f(a) and f(b) have opposite signs [i.e., f(a)f(b) < 0]. Then, there is at least one number c (a,b) for which f(c) = 0. Remark: The intermediate value property has many applications. In particular, it can be used to estimate a solution of a given equation. 68/69

Example 16 Show that the equation x 2 x 1 = 1 has a solution between x +1 1 < x < 2. Proof: Let f(x) = x 2 x 1 1. Then f(1) = 3/2 and x +1 f(2) = 2/3. Since f(x) is continuous for 1 x 2, it follows from the intermediate value theorem that the curve must cross the x axis somewhere between x = 1 and x = 2. In other words, there is a number c such that 1 < c < 2 and f(c) = 0, i.e., c 2 c 1 = 1 c +1. 69/69