3.5 Continuity of a Function One Sided Continuity Intermediate Value Theorem... 23

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1 Chapter 3 Limit and Continuity Contents 3. Definition of Limit Basic Limit Theorems One sided Limit Infinite Limit, Limit at infinity and Asymptotes Infinite Limit and Vertical Asymptote Limit at Infinity and Horizontal Asymptotes Continuity of a Function One Sided Continuity Intermediate Value Theorem Eercise 25 The development of calculus was stimulated by two geometric problems: finding areas of plane regions and finding tangent lines to curves. Both of those problems require a it process for their general solution. However, it process occurs in many other applications as well. Besides the concept of it is the fundamental building block on which all other calculus concepts are based. 3. Definition of Limit Limits described what happens to a function f () as its variable approaches to a particular number a but not on a. Definition 3.. Let f be a function defined in an open interval containing a, with the possible eception of a it self. Then, the it of the function at a is the number L, written as f () = L a Eample 3. Let f () = + 4, what happens to f () as approaches to, but not equal to one. Solution: To investigate the behavior of f () as approaches numerically and graphically we can construct a table and draw a graph of f () for near. The above table is as approaches from the right The above table is as approaches from the left. From the above table, we can conclude that f () approaches 5 as approaches from both the left and right side of f() Any one can get this soft-copy from Google site Eodus4Wisdom

2 2 Limit and Continuity f() Figure 3.: f () = + 4 Eample 3.2 Find f () for a function f () = 2. 2 Solution: The following table illustrates the behavior of the function, as becomes closer and closer to 2 from both the left and right side of 2. The above table is as approaches 2 from the right; that is, f () The above table is as approaches 2 from the left; that is, f () Therefore, the value of f () = 2 are near 4 whenever is close to 2 from both the left and right sides Figure 3.2: f () = 2 since f () = f () = f () = 4. Definition 3..2 Let f be a function defined in an open interval containing a, with the possible eception of a it self. Then, the it of the function at a is the number L, written as for all number ε > 0, there eist a number δ > 0 such that Eample 3.3 Evaluate a. Solution: Let ε > 0 be any number. Take δ = ε. Now 0 < a < δ f () L < ε a < δ a < ε f () a < ε since f () = Hence, a = a f() Any one can get this soft-copy from Google site Eodus4Wisdom

3 3. Definition of Limit f() Figure 3.3: The it of f on the open interval containing a Eample 3.4 Show that 5 f () = 4 by using ε δ definition, where f () = 2 6. Proof: Let ε > 0 be given. We must find δ > 0 such that Consider Now choose δ = ε 2. Thus, 5 < δ f () 4 < ε f () 4 < ε (2 6) 4 < ε 2 0 < ε 5 < ε 2 Therefore, 5 < δ f () 4 < ε. 0 < 5 < δ 5 < ε < ε 2 0 < ε (2 6) 4 < ε Eample 3.5 Given f () = 4, find δ > 0 for a number ε = such that if 0 < a < δ, 2 then f () L < ε. Solution: For every ε = > 0 such that So that, 0 < ( 2) < δ f () ( 4) < ε 0 < ( 2) < δ (5 + 6) + 4 < ( + 2) < < = Thus, we can choose δ = 0.00 or any positive number less than Eample 3.6 Show that 4 f () = 0 by using ε δ definition, where f () = 2 6. Proof: Let ε > 0 be given. We must find δ > 0 such that 4 < δ f () 0 < ε Any one can get this soft-copy from Google site Eodus4Wisdom

4 4 Limit and Continuity Consider f () 0 < ε ( 2 6) 0 < ε 2 6 < ε ( + 4)( 4) < ε Let choose δ =. Thus, 0 < 4 < < 4 < 3 < < 5 7 < + 4 < < < 9 4 Let for some values of we have 9 4 < ε 4 < ε 9. So choose δ 2 = ε 9. Now choose δ = min{δ,δ 2 } = {, ε 9 }. 0 < 4 < δ = ε 9 Therefore, 4 < δ f () 0 < ε. 4 < ε < ε < ε since 4 < 9 ( + 4)( 4) < ε 2 6 < ε ( 2 6) 0 < ε f () 0 < ε Eample 3.7 Show that f () = by using ε δ definition, where f () = +2. Proof: Let ε > 0 be given. We must find δ > 0 such that 4 < δ f () 2 3 < ε Consider Let choose δ =. Then, f () 2 3 < ε 2 2( + 2) 3( + 2) < ε 8 2 3( + 2) < ε < ε < ε 4 < δ = < 4 < 3 < < 5 5 < + 2 < 7 5 < + 2 < 7 7 < + 2 < < < Any one can get this soft-copy from Google site Eodus4Wisdom

5 3. Definition of Limit 5 Let for some values of we have < ε 4 < 5 2 ε. So choose δ 2 = 5 2 ε. Now choose δ = min{δ,δ 2 } = {, 5 2 ε}. 0 < 4 < δ = ε 4 < 2 ε 2 4 < ε < ε since + 2 < < ε 8 2) 3( + 2) < ε 2 2( + 2) 3( + 2) < ε f () 2 3 < ε Therefore, 4 < δ f () 2 3 < ε. Eample 3.8 Show that f () = 5 by using ε δ definition, where f () = 4 +. Proof: Let ε > 0 be given. We must find δ > 0 such that Consider < δ f () 5 < ε f () 5 < ε < ε < ε < ε + Let choose δ =. Then, < δ = < < 0 < < 2 0 < < 2 + < ε < + < 2 + < < < 2 + < = + Let for some values of we have < ε. So choose δ 2 = ε. Now choose δ = min{δ,δ 2 } = {,ε}. 0 < < δ < ε < ε since + < = + < ε + < ε < ε f () 5 < ε Any one can get this soft-copy from Google site Eodus4Wisdom

6 6 Limit and Continuity Therefore, 4 < δ f () 2 3 < ε. 3.2 Basic Limit Theorems If f () = L and g() = M, then a a. Limit of a constant function; that is, f () = c for all for any constant c c = c a This means the graph of the constant function f is a horizontal line. That is no matter what the value of, f () is always c. Thus as we approach = c from either the left or the right, we hit the line y = c at height of c. Eample = 5 2. Limit of identity function; that is, f () = for all = a a The graph of the function is a straight line. As we approach the point = a from the left and the right, the function approaches the value a. Eample 3.0 = Constant Multiple Rule c f () a = c a = cl for any constant c Eample 3. 3 = 3 = 3(2) = Sum Difference Rule ( f () ± g()) a = f () ± a a = L ± M Eample 3.2 ( + 6) = + 6 = 3(2) + 6 = Product Rule ( f () g()) a = f () a a = L M Eample = 2 = 2 = 2 2 = 4 2 Eample 3.4 ( + 2 )(2 2) = ( + ) ( 2 2) 2 2 = 3 2 = 6 Any one can get this soft-copy from Google site Eodus4Wisdom

7 3.2 Basic Limit Theorems 7 6. Quotient Rule f () a g() = f () a g() a = L provided that M 0 M 7. Power Rule a ( f ())n = ( a f ()) n = L n, n is any real number Eample 3.5 (3 + 2) 2 = ( 3 + 2) 2 = 8 2 = Although the above basic it theorems are stated for only two functions f and g, the result will still be true for a finite number of functions. Eample 3.6 Find the ( ). Solution: ( ) = = 5( ) = 5() 2 + 3() + = 9 Eample 3.7 Find the Solution: = ( ) = 2 2 7( ) (2) = 7(2) 2 + 2(2) + = From the above eamples, we can have the following formulas to evaluate its of a polynomial and rational functions. If f () and g() are polynomials, then f () f () = f (a) and a a g() = f (a) g(a),g(a) 0 In both cases you simply substitute the value of a in to the equation. Eample 3.8 Find the +3. Solution: Since ( ) = 0, the quotient rule for its does not apply here. When the denominator of the given rational function approaches zero, while the numerator does not, we can conclude that the it does not eist. When both the numerator and the denominator of the given rational function approaches zero, simplify the function algebraically in order to find the desired its. The its of f () as approaches a, depends on the values of f () as becomes close to a, but we eclude = a. Any one can get this soft-copy from Google site Eodus4Wisdom

8 8 Limit and Continuity Eample 3.9 Evaluate Solution: When substitute 2 in to the function, then both the numerator and denominator becomes zero, hence f (2) is meaningless. Therefore, straight substitution does not yield the it of the function f at = 2. Simplifying the function, we get Thus, f () = = ( 2)( + ) ( 2) 2 2 = ( + ) = = ( + ) Remark 3.2. Even after epressing f in the form of +, we do not think what + is when = 2 but rather that of what + approaches as tends to 2. Eample 3.20 Evaluate Solution: Observe that 2 +3 is the quotient of 2 and + 3, whose its at 2. We know that We conclude from the quotient rule that 2 2 = 4 and ( + 3) = = = ( + 3) Eample 3.2 Show that does not eist. Solution: Suppose that eists, and let L =. Since = ( ), by using product rule = = ( ) = = 0 L = 0 Which is obvious false. Therefore, does not eist. Eample 3.22 Find Solution: Since 2 (2 4), we can not apply the quotient rule to this function in its original form. However, since = ( + 2)( 2 ) and 2 4 = ( + 2)( 2), we have = ( + 2 2)(2 ) ( + 2)( 2) 2 = 2 2 = ( 2)2 2 2 = 3 4 Any one can get this soft-copy from Google site Eodus4Wisdom

9 3.2 Basic Limit Theorems 9 Eample 3.23 Evaluate ( 3) 9 9. Since 9 ( 9) = 0, the quotient rule can not be applied directly. However, if we factor the denominator, then we can cancel terms and the quotient rule ( 3) 9 9 ( 3) = 9 ( 3)( + 3) = = = 3 2 Theorem Squeezing Principle Suppose that f () h() g() for all a in some neighborhood. Suppose also that Then, we also have a h() = k. f () = k = g() a a Theorem Important Limit Theorems sin sin cos 2 = 2 ( + ) = e e cos = tan = = 0 +( + ) = e = ln = Eample 3.24 Show that = Solution: Construct a circle with center at O and radius OA = OD =, as in figure below. Choose point B on A etended and point C on OD so that lines BD and AC are perpendicular to OD. It is geometrically Figure 3.4: Circle with center O evident that Area of OAC < Area of a sector OAD < Area of OBD (3.) Any one can get this soft-copy from Google site Eodus4Wisdom

10 0 Limit and Continuity From this 2 sincos < 2 < tan (3.2) 2 Divide equation (3.2) both sides by 2 sin, then we have cos < sin < cos cos < sin < cos As 0, cos and cos, by squeezing theorem it follows that sin = sink tank Remark = k and = k for any constant k. sin n = for any positive integer n. Eample 3.25 Show that cos = 0 Solution: Notice that = 0, so we can not apply the quotient rule directly. However cos = ( cos ) ( cos + cos + ) cos 2 = (cos + ) sin 2 = (cos + ) = ( sin sin )( cos + ) Since =. Furthermore, + = 0. By the sum and the quotient rules. Thus the product rule tells us that sin sin cos+ = 0 cos = ( sin sin )( cos + = 0 = 0 Eample 3.26 Evaluate 2. Any one can get this soft-copy from Google site Eodus4Wisdom

11 3.2 Basic Limit Theorems Solution: 2 ( 2 ) + = = = ( 2 ) + ( ) = ( )( + ) + ( ) = ( )( ( + ) + ) = = ( ( + ) + ) = 3 Eample 3.27 Is there a number a such that a + a eist? If so find the value of a and the value of the it. 3 Solution: Since the it of 2 +a+a+3 eist, the numerator must be the factor of ( + 2); that is, Thus So, a + a a + a = ( )( + 2) a + a + 3 = (b + c)( + 2) a + a + 3 = b 2 + (2b + c) + 2c b = 3,2b + c = a and a + 3 = 2c a = 5,b = 3 and c = a + a (3 + 9)( + 2) = 2 ( )( 2) = 2 = Eample 3.28 Show that 3 sin = 0. Solution: First notice that we can not use 3 sin = 3 sin because sin However, since does not eist. sin Any one can get this soft-copy from Google site Eodus4Wisdom

12 2 Limit and Continuity We have, 3 3 sin 3. We know that 3 = 0 and ( ) 3 = 0 Taking f () = 3, g() = 3 sin, and h() = 3 in the squeezing theorem, we obtain 3 sin = One sided Limit Definition Let f be a function which is defined at every number in some open interval (a,c). Then the it of f (), as approaches a from the right, is L, written a + f () = L if for any ε > 0, however small, there eist a δ such that, f (( L < ε whenever 0 < a < δ. 2. Let f be a function which is defined at every number in some open interval (d,a). Then the it of f (), as approaches a from the left, is L, written f () = L a if for any ε > 0, however small, there eist a δ such that, f (( L < ε whenever δ < a < The it of a function eists at = a if and only if f () = L = f () a a + Eample 3.29 Find the it of f () as approaches 2 from the left and the right, where f () = 3 and determine whether the it of the function eist at = 2 or not. Solution: First we have to evaluate one sided its from both sides of = 2 f () 2 = f () 2 + = Therefore, f () =. 2 { +, < Eample 3.30 Find f (), where f () = 3, > Solution: Evaluate the it as approaches. From the graph we can see that = 2 + = 2 Therefore, f () = 2. Eample 3.3 Find f (), where f () = {. {, 0, > 0 Solution: Notice that = and, < 0 =, < 0 Then the left side it can be found as f () = = = Any one can get this soft-copy from Google site Eodus4Wisdom

13 3.4 Infinite Limit, Limit at infinity and Asymptotes 3 Figure 3.5: f () = + if < and f () = 3 if > and the right side it can be found as f () = + + = = + Therefore, f () does not eist since the left side and right side its are not the same; that is, f () f (). Graphically, it is shown in the figure below. The value of + approach different Figure 3.6: f () = numbers as approaches 0 from different sides, so does not eist. 3.4 Infinite Limit, Limit at infinity and Asymptotes 3.4. Infinite Limit and Vertical Asymptote Definition 3.4. Let f be a function defined in an interval containing a, with the possible eception of a it self. Then,. f () =, if for every number M > 0 there is some δ > 0 such that 0 < a < δ, then a + f () > M. 2. f () =, if for every number M < 0 there is some δ > 0 such that 0 < a < δ, then a + f () < M. 3. f () =, if for every number M > 0 there is some δ > 0 such that δ < a < 0, then a f () > M. 4. f () =, if for every number M < 0 there is some δ > 0 such that δ < a < 0, then a f () < M. Eample 3.32 Let f () =, show that f () = and f () =. + + Any one can get this soft-copy from Google site Eodus4Wisdom

14 4 Limit and Continuity Solution: For every M > 0, we can find δ > 0 such that 0 < 0 < δ f () > M 0 < < δ > M < M Now choose δ = M. Thus 0 < 0 < δ > δ = M f () > M Therefore, f () = and for every M < 0, we can find δ > 0 such that + δ < 0 < 0 f () < M δ < < 0 < M > M Now choose δ = M. Thus δ < 0 < 0 < δ = M f () < M Therefore, f () =. 5+ Eample 3.33 Show that f () =, where f () = 2. 2 Solution: For every M < 0, there is a number δ > 0 such that δ < 2 < < M Consider < M > M 2( 2 ) 5 + > M Any one can get this soft-copy from Google site Eodus4Wisdom

15 3.4 Infinite Limit, Limit at infinity and Asymptotes 5 Let choose δ 5, then Thus, 5 < δ < 2 < 0 5 < 2 < 0 5 < 2 < < < < 5 < < 5 + < < 5 + < < < ( 2 ) < ( 2 ) < 4 7 ( 2 ) since 2 < 0 2( 2 ) 5 + > M M < 2( 2 ) 5 + < 4 7 ( 2 ) 5 + M < 4 7 ( 2 ) 5 + M < M < 2 Choose δ 2 = 4M 7. Now choose δ = min{δ,δ 2 } = { 5, 4M 7 }. Thus, δ < 2 < 0 7 4M < 2 < 0 7 4M < 2 7 2M < < M (5 + ) 2 < < M f () < M 2 < M Therefore, f () =. 2 Definition The line = a is called a vertical asymptote of the graph of y = f () if any one of the following its holds true f () = ± a Any one can get this soft-copy from Google site Eodus4Wisdom

16 6 Limit and Continuity f () = ± a + f () = ± a Eample 3.34 The line = 0 is a vertical asymptote of the graph f () = + =. since = and Figure 3.7: y = Definition The point = 0 is a hole to the graph of a rational function f if and only 0 f () eist and f ( 0 ) does not defined. Eample 3.35 Determine the vertical asymptote or a hole to the graph of f () = 2, if it eists. 2 Solution: The zeros of 2 = 0 are = 0 or =. So, f is not defined at = 0 and =. Now, find the it of f () at = 0 and =. 2 2 = + = 2 Thus, f () has a hole at = and Thus, f () has a vertical asymptote at = = ± Limit at Infinity and Horizontal Asymptotes Definition Let f be a function defined in an interval (a, ), then f () = L if and only if for all ε > 0, there eist M > 0 such that if > M, then f () L < ε. 2. Let f be a function defined in an interval (,a), then f () = L if and only if for all ε > 0, there eist N < 0 such that if < N, then f () L < ε. Definition The line y = L is called a horizontal asymptote of the graph of y = f () if either f () = L or f () = L. Eample 3.36 Consider the behavior of the function f () =, 0 when tends to infinity. Solution: As gets larger and larger and continuous to grow without bound, the corresponding values of f get closer and closer to 0 and eventually tend to 0. The values of f () becomes closer and closer to zero as approaches to infinity f () = 0 Any one can get this soft-copy from Google site Eodus4Wisdom

17 3.4 Infinite Limit, Limit at infinity and Asymptotes 7 This is read as the it of f is zero as tends to positive infinitive. Similarly the it of f is zero as tends to negative infinitive. f () = 0 Since = 0 and = 0, the line y = 0 is the horizontal asymptote of f () =. Figure 3.8: y = Eample 3.37 Evaluate Solution: Let us divide both the numerator and denominator by Since we are concerned with the behavior of = 3 + 2, 0, 2 for sufficiently large values of. Thus we have = = Therefore, +2 = 3. Eample 3.38 Find the vertical and horizontal asymptotes of the graph of f () = 3. Solution: To find Vertical Asymptotes Notice that if f () a = ± f () a + = ± or a f () = ± Then the line = a is a vertical asymptote of f (). Now, find = and also =. Thus, the line = 3 is the vertical asymptote To find Horizontal Asymptotes If either f () = L or f () = L, then the line y = L is called a horizontal asymptote of the graph of y = f (). Now, find = 0 is horizontal asymptote. ± 3 Any one can get this soft-copy from Google site Eodus4Wisdom

18 8 Limit and Continuity Theorem 3.4. The it of a function (if eists) is unique. 3.5 Continuity of a Function Definition 3.5. A function f is said to be continuous at a if a f () = f (a) Definition If f is continuous at a, then. f (a) is defined 2. a f () eist 3. a f () = f (a). If f is not continuous at a, then we say that f is discontinuous at a. Figure 3.9: Continuity of f () as approaches a If a is in the domain of the rational function f, then a f () = f (a). Thus any rational function is continuous at every point in its domain. Eample 3.39 Show that the polynomial function f () = is continuous at =. Solution: First we have to check whether the above three condition of continuity are satisfied or not. f () is defined 2. f () eist 3. f () = f () Therefore, the function f () = is continuous at =. Eample 3.40 Show that f () = 2 is not continuous at =. Solution:. f () is not defined 2. f () eist Even if the it of the function eists, since it is not defined at =, the function f () is discontinuous at =. Eample 3.4 Find the points of discontinuity of f () = Solution: The three condition for continuity are satisfied for any values of R ecept at = 2. Therefore, = 2 is the only discontinuity point for f () = One Sided Continuity Any one can get this soft-copy from Google site Eodus4Wisdom

19 3.5 Continuity of a Function 9 Definition f is said to be continuous from the right at a if f () = f (a). a + 2. f is said to be continuous from the left at a if f () = f (a). a Figure 3.0: Continuity of f as approaches to a from the right Figure 3.: Continuity of f as approaches to a from the left Eample 3.42 Show that f is continuous from the right at 0, but not continuous from the left at 0, where f () defined by { if < 0 f () = if 0 Solution: Since f () = = and f (0) =, it follows that f () = f (0) Hence f is continuous from the right at zero. On the other hand, f () = = and f (0) =. since f () f (0), f is not continuous from the left at zero. Figure 3.2: Eample 3.43 Show that the function defined by { 0 if < 0 f () = if 0 Any one can get this soft-copy from Google site Eodus4Wisdom

20 20 Limit and Continuity is continuous from the right at 0 but not continuous from the left at 0. Solution: f () = = f (0) because of this f is continuous from the right at 0, f is not continuous + from the left at 0 since f () = 0 f (0). Definition A function f is said to be. Continuous on (a,b) if f is continuous at each point in the open interval (a,b). 2. Continuous function if f is continuous over its domain. 3. Continuous in the closed interval [a,b] if f is continuous at each point in the open interval (a,b) and = a from the right and at = b from the left. Eample 3.44 Show that the function f () = 2 +5 is continuous on the open interval ( 3,2). Solution: The three conditions for continuity are satisfied for for any value of between -3 and 2. Therefore, the function is continuous in the open interval ( 3,2). However, the function is not continuous for the closed interval [ 3,2], since f () is discontinuous at = 2. Eample 3.45 Show that f () = is continuous on [0,2]. Solution: a = a for every a > 0, it follows that f is continuous at a. Hence f is continuous on (0,2). Moreover, 2 = 2 and f (2) = 2. Thus f is continuous from the left at 2. Similarly, + = 0 = f (0) and so f is continuous from the right at 0. Hence f is continuous on [0,2]. Eample 3.46 Let f () = Determine the numbers at which f is continuous Solution: Observe that f is a rational function. The denominator is 0 for = and = 6, so f is defined for all ecept and -6. Therefore f is continuous at every number ecept and -6. Theorem 3.5. A function is continuous at a if and only if it is both continuous from the right and continuous from the left at a. Eample 3.47 Determine the value of a constant a such that the function { a f () = if 3 2a + if < 3 is continuous at = 3. Solution: If f () is to be continuous at = 3, then 3 f () must eist and furthermore f () = f (3) 3 f () 3 + = + 2) = 9a + 2 and 3 +(a2 f () = + ) = 6a + +(2a 3 3 Now equate the left and right side its. This means 3 f () = f () = f () = f (3) + 3 From this it follows that 9a + 2 = 6a +. Solving for a we get a = 3. Therefore the value of a must be 3 if f () is to be continuous at = 3. 3 Any one can get this soft-copy from Google site Eodus4Wisdom

21 3.5 Continuity of a Function 2 Theorem If f and g are continuous at a and c is a constant, then f ±g, c f, f g and f g, if g(a) 0 are also continuous at a. Theorem If f () is continuous at = a, then f () is also continuous at = a. Eample 3.48 f () = is continuous at =, then is also continuous at =. Figure 3.3: f () = Theorem Polynomial functions, Rational functions, Root functions, Trigonometric functions, Inverse functions, Eponential functions and Logarithmic functions are continuous on their domain. Theorem If a g() = b and f is continuous at b, then a f (g()) = f (b); that is, f (g()) = f ( g()) a a Theorem If g is continuous at a and f is continuous at g(a), then f og is continuous at a; that is, f (g()) = f (g(a)) a Eample 3.49 Show that h is continuous at 2, where h() =. Solution: Let g() = and f (y) = y. Then h = f og. We know that g is continuous at = 2 and that f is continuous at g(2) =. Since the square root function is continuous at every positive number. It follows from the above theorem that it is continuous at Intermediate Value Theorem Theorem Suppose f is continuous on a closed interval [a,b]. Let K be any number between f (a) and f (b), so that f (a) k f (b) or f (b) k f (a). Then there eist a number c in [a,b] such that f (c) = K. Notice that f be continuous on an interval I. If f has both positive and negative values on I, then the intermediate value theorem implies that f () = 0 for some in I; that is, f has zero in I. Equivalently, if f has no zero in I, then either f () > 0 for all in I or f () < 0 for all in I. Theorem Let f be continuous on [a,b] and f (a) < 0 < f (b) or f (b) < 0 < f (a), then the function have at least one solution. Eample 3.50 Show that the equation has solution between 0 and 2. Solution: Let f () = Any one can get this soft-copy from Google site Eodus4Wisdom

22 22 Limit and Continuity since f is a polynomial function, then f is continuous on R. Here f (0) = 3 and f (2) = 7. Thus, f (0) < 0 < f (2); that is, 0 is between f (0) and f (2). So, by Intermediate Value Theorem there is a number c between 0 and 2 such that f (c) = 0. Eample 3.5 Let f () = 3. Find the solution of f on [ 2,2]. Solution: Here, f ( 2) = 6 and f (2) = 6, because of this f ( 2) < 0 < f (2). By intermediate value theorem, there eist c in [ 2,2] such that f (c) = 0. So, c is the solution of f. f (c) = c 3 c = 0 c(c 2 ) = 0 Thus,, 0 and are the solution of f () = 3. c(c )(c + ) = 0 c = 0,c = or c = Eample 3.52 If f () = 2 2 +, show that there is a number c such that f (c) = 0. Solution: The function f () is continuous every where, so we can use the intermediate value theorem. Let take the interval [0,3] R, then at = 0, f (0) = (0) 3 (0) 2 +0 = 0 and = 3, f (3) = (3) 3 (3) 2 +3 = 2. Since 0 f (c) , by Intermediate value theorem, there eist 0 c 2, such that f (c) = 0. Theorem Every continuous function in a closed interval [a, b] attains its maimum and minimum value. Any one can get this soft-copy from Google site Eodus4Wisdom

23 3.6 Eercise Eercise. For the function g whose graph is given, state the value of each quantity, if it eist. If it does nor eist, eplain why. (a) g() (d) g() 2 (g)g(2) (b) g() + (e) g() 2 + (h) g() 4 (c) g() ( f ) g() 2 Figure 3.4: graphs of g() Ans., 2, does not eist, 2, 0, does not eist,, 3 2. Evaluate the infinite it Evaluate the infinite it (+2). Ans. Ans. 4. Given that f () = 4, g() = 2, h() = Find the its that eist. If the its does not eist, eplain why. (a) 2 [ f () + g()] (d) 2 3 f () g() (b) 2 [g()] 3 (e) 2 + g() h() (c) 2 f () ( f ) 2 g()h() h() 5. Evaluate the it ( ) 3. Ans. 6, 8,2, 6,does not eist,0 6. Evaluate the it ( ) 3. Ans Evaluate the it Evaluate the it 4 ( ). Ans. 8 Ans. 5 Any one can get this soft-copy from Google site Eodus4Wisdom

24 24 Limit and Continuity Ans Let f () = 2. (a) Find f () and f (). + (b) Does f () eist? f () 8 0. If = 0, find f (). Ans. 2, 2, No; does not eist Ans. 8. For the it ( ) = 2 illustrate the definition it by finding values of δ that corresponding to ε = and ε = 0.. Ans. 0. and 0.02(or smaller positive numbers) 2. Prove that ( ) = by using ε δ definition of it Show that f () = ( ) 4 is continuous at a =. 4. Show that f () = 2 3 is continuous on the interval (2, ). 5. Eplain why the function f () = 4 +7 is continuous at every number in its domain. State the domain. { 2 if < 6. Show that f () is containing on (, ), where f () = if { sin if < π 7. Show that f () is containing on (, ), where f () = 4 cos if π 4 8. Show that the function { f () = 4 sin( 4 ) if 0 0 if = 0 Ans. { 2, 3 } is continuous on (, ). 9. Use the intermediate value theorem to show that there is a root of the equation = 0 in the interval (,2). 20. Use the intermediate value theorem to show that there is a root of the equation 3 = in the interval (0,). Any one can get this soft-copy from Google site Eodus4Wisdom