Unit 1: Limits and Continuity of a Function 1.1 LIMITS AND CONTINUITY OF A FUNCTION U1 D i f f e r e n t i a t i o n
Unit 1: Limits and Continuity of a Function 1.2
Unit 1: Limits and Continuity of a Function 1.3 Study organiser Before you begin this unit, please check through your study organiser. It shows the topics that you’ll be covering, the skills you need to acquire (the learning outcomes) and the activities you’ll do to help you acquire these skills. T o p i c L e a r n i n g O u t c o m e s K e y C o n c e p t s A c t i v i t y 1.1 Limit of a Function Evaluate a limit numerically, graphically and algebraically. • 2x − → , means x approaches 2 from the left. • 2x + → , means x approaches 2 from the right. • A limit will exist if the left hand limit equals the right hand limit. • When given a graph a limit will exist when: 1. the graph passes through a point. 2. two pieces of the graph meet at a point where there is a circle. • When given a graph a limit will not exist when there is a: 1. vertical asymptote. 2. break or step in the graph. • A circle means the graph does not pass through that point. • A dot means the graph passes through that point. • Given a rational function, we factorise where possible, cancel any like factors and then substitute the given x − value to find the limit. • Given an expression with a square root, we rationalize the expression, cancel like factors and then substitute the given x − value to find the limit. • When given a piecewise function, we determine the left hand limit and right hand limit to calculate the limit of the function. Ex. 1.1
Unit 1: Limits and Continuity of a Function 1.4 T o p i c L e a r n i n g O u t c o m e s K e y C o n c e p t s A c t i v i t y 1.2 Continuity of a Function To be able to determine if a function is continuous or discontinuous at certain x − values. • A function ( )f x is continuous at the value x c= if: 1. ( )f c is defined. 2. lim ( )f x exists. 3. lim ( ) ( )f x f c= . • A function shown by a graph is discontinuous where there is a: 1. hole shown by a circle. 2. vertical asymptote. 3. jump. • A rational function is discontinuous for x − values that make the denominator zero. Ex. 1.2 x c→ x c→ S t u d y t i m e It is suggested that you spend at least 4 hours on this section.
Unit 1: Limits and Continuity of a Function 1.5 1.1 Limit of a Function The notion of limits is a fundamental concept of calculus. In this course, you will learn how to evaluate limits and how they are used in the three basic problems of calculus: the instantaneous rate of change, gradient of tangent line and to find the derivative. To evaluate limits, we will look at three methods: estimating limit numerically, graphically and algebraically. D e f i n i t i o n o f L i m i t This definition of limit examines the behavior of the function ( )f x when the value of x is close to the value a . It is not concerned with the value of ( )f a . In fact, the value of ( )f a may even be undefined. As long as the value of ( )f x approaches a unique real number as x a→ , the limit exists. This idea will be shown in the second example. E s t i m a t i n g a L i m i t N u m e r i c a l l y Use a table to estimate the limit numerically for 2 2 lim ( 3 4). x x x → + − Let 2 ( ) 3 4f x x x= + − . Therefore, 2 (2) 2 3(2) 4f = + − 4 6 4 6= + − = Let us examine the behaviour of this function close to 2x = . The number 2 can be approached from two sides. This is shown by the diagram below. x approaches 2 from the left 2 x approaches 2 from the right Construct a table that shows values of ( )f x for two sets of x -values; one set that approaches 2 from the left and one that approaches 2 from the right side. x 1.9 1.99 1.999 2 2.001 2.01 2.1 ( )f x 5.31 5.9301 5.993 6.007 6.0701 6.71 If ( )f x becomes arbitrarily close to a unique number L as x approaches c from either side, then the limit of ( )f x as x approaches to a is L . This is written as: lim ( ) x a f x L → = Example 1.1.1
Unit 1: Limits and Continuity of a Function 1.6 From the table, it appears that the closer x gets to 2, the closer ( )f x gets to 6. So, you can estimate the limit to be 6. We can see that: 1. As 2x − → , ( ) 6f x → , i.e. 2 lim ( ) 6 x f x− → = . This is the left sided limit. 2. As 2x + → , ( ) 6f x → , i.e. 2 lim ( ) 6 x f x+ → = . This is the right sided limit. Therefore, the two-sided limit 2 lim ( ) 6 x f x → = exists only if both one-sided limits are the same. Use a table to estimate the limit numerically for 2 3 4 3 lim . 3x x x x→ − + − You should be able to see that (3)f is undefined. 2 3 4(3) 3 0 (3) 3 3 0 f − + = = = − not defined, we cannot divide by zero. Let us examine the behaviour of ( )f x as the value of x approaches 3. x 2.5 2.9 2.99 2.999 3 3.001 3.01 3.1 3.5 ( )f x 1.5 1.9 1.99 1.999 2.001 2.01 2.1 2.5 Example 1.1.2 NOTE 1. The phrase “ x approaches 2 from the left” is shown by 2x − → . The negative sign means from the left side of 2. 2. The phrase “ x approaches 2 from the right side” is shown by 2x + → . The positive sign means from the right side of 2. 3. However, the phrase “ x approaches 2” is shown by 2x → . You must understand the notation in 1 and 2 above, as it will be used in later examples. 4. Limit will only exist if the left side limit is equal to the right side limit. lim ( ) . lim ( )f x f x x a x a =− +→ →
Unit 1: Limits and Continuity of a Function 1.7 We can see that: 1. As 3x − → , ( ) 2f x → , i.e. 3 lim ( ) 2 x f x− → = . 2. As 3x + → , ( ) 2f x → , i.e. 3 lim ( ) 2 x f x+ → = . Since ( )f x approaches the same value as x approaches 3 from the left and right, we can conclude that 3 lim ( ) 2 x f x → = . Therefore, this shows that ( )f x has a limit as 3x → even though the function is not defined at 3.x = This often happens and it is important to realize that the existence or nonexistence of ( )f x at x a= has no bearing on the existence of the limit of ( )f x as x approaches a . You should therefore understand from the previous example that the left-sided limit and the right-sided limit must be equal for the limit to exist. Suppose we had a function ( )f x that produced the table of values shown below. What would we say concerning the value of the 2 ( )lim x f x → . x 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5 ( )f x 3.75 4.31 4.9301 4.993 6.007 6.0701 6.71 9.75 We can see that: 1. As 2x − → , ( ) 5f x → , i.e. 2 lim ( ) 5 x f x− → = . 2. As 2x + → , ( ) 6f x → , i.e. 2 lim ( ) 6 x f x+ → = . Since ( )f x approaches a different value as x approaches 2 from the left and right side, we can conclude that: 2 lim ( ) x f x → = Does not exist Example 1.1.3 The two sided limit of a function ( )f x exists at a point a if and only if the one-sided limits exist at that point and have the same value; that is, 1( )lim x a f x L− → = and 2( )lim x a f x L+ → = , Then ( )lim x a f x → exists only if 1 2L L= .
Unit 1: Limits and Continuity of a Function 1.8 Compute the 2 1 1 1 lim x x x→ + − , if it exists. For this function, (1)f is undefined, since we have zero in the denominator. The table below shows calculated values for ( )f x as the value of x approaches 1. x 0.9 0.99 0.999 1 1.001 1.01 1.1 ( )f x 18.1− 198.1− 1,998.1− 2,002.001 202.10 22.1 From the table, it appears that as 1x + → , the value of ( )f x gets larger and larger in the positive direction and as 1x − → , the value of ( )f x gets larger and larger in the negative direction. That is, the value of ( )f x does not approach a unique real number as 1x → . Therefore, we say that 2 1 1 1 lim x x x→ + − does not exist. We can also arrive at the same conclusion if we look at the graph of the function 2 1 ( ) . 1 x f x x + = − ... As 1x − → the value of ( )f x → − ∞ and as 1x + → the value of ( )f x → + ∞ therefore, 2 1 1 lim 1x x x→ + − does not exist. Example 1.1.4
Unit 1: Limits and Continuity of a Function 1.9 U s i n g a G r a p h t o F i n d t h e L i m i t Find the limit of ( )f x as x approaches 1, where ( )f x is defined as 2, 1 ( ) . 2 , 1 x x f x x x ≤ = > These types of functions are called piecewise functions. They can also have three pieces. You should note that ( )f x is defined at 1x = . To calculate its value, we use the piece with the equals sign. 2 ( ) , 1f x x x= ≤ Therefore, we have: 2 (1) 1 1f = = Let us look at the graph of the function ( )y f x= to find the limit. We can see that: 1. As 1x − → , ( ) 1f x → , i.e. 1 lim ( ) 1 x f x− → = . 2. As 1x + → , ( ) 2f x → , i.e. 1 lim ( ) 2 x f x+ → = . Since the value of ( )f x does not approach a unique real number as 1x → , we conclude that 1 lim ( ) x f x → does not exist. Example 1.1.5
Unit 1: Limits and Continuity of a Function 1.10 We will now look at the graph of a function ( )f x below and determine the value of certain limits. Evaluate the following. a. ( 3)f − This means we find the y value when 3x = − . From the graph you should be able to see that when 3x = − , 2y = Therefore, we have: ( 3) 2.f − = b. ( 1)f − At 1x = − , there is a vertical asymptote, which means that the function is undefined at that x value. Therefore, we have ( 1)f − = undefined. Example 1.1.6 NOTE: Conditions under which limits do not exist The does not exist under any of the following conditions. 1. The value of becomes infinitely large or infinitely Example 4 small as the value of approaches from either side. 2. and where . Example 3, 5
Unit 1: Limits and Continuity of a Function 1.11 c. (1)f At 1,x = there is a DOT at the point (1, 2) and a HOLE at the point (1,3) . We consider the DOT as the graph can pass through this point. Thus, when 1x = , 2y = Therefore, (1) 2.f = d. (4)f At 4x = , there is a HOLE at the point (4, 5) . This means that the graph does not pass through this point, so the function is UNDEFINED at 4x = . e. 3 lim ( ) x f x− →− For this type of question we approach 3− from the left and see to what y value does the graph approaches. You should be able to see that the straight line approaches the value 2.y = Therefore, we have: 3 ( ) 2lim x f x− →− = f. 3 ( )lim x f x+ → − As x approaches 3− from the right side, the value of ( )f x approaches 2. Therefore 3 ( ) 2lim x f x+ →− = g. 3 ( )lim x f x →− Since 3 ( )lim x f x− → − = 3 ( ) 2lim x f x+ →− = , then, 3 2lim ( ) x f x →− = . h. 1 ( )lim x f x → − As x approaches 1− from the right side or the left side, the value of ( )f x increases without bound. Note that at 1x = − we have a vertical asymptote. Because ( )f x is not approaching a unique real number L as x approaches 1− , you can conclude that limit does not exist. That is 1 ( )lim x f x →− does not exist. NOTE When two different pieces of graph MEET AT A POINT, the limit exists for the given value and is equal to the value of the point.
Unit 1: Limits and Continuity of a Function 1.12 i. 1 ( )lim x f x− → As x approaches 1 from the left side, the value of the function approaches 3. Therefore, 1 ( ) 3lim x f x−→ = j. 1 ( )lim x f x+→ As x approaches 1 from the right side, the value of the function approaches 3. Therefore, 1 ( ) 2lim x f x+ → = k. 1 ( )lim x f x → Since 1 ( ) 3lim x f x− → = and 1 ( ) 2lim x f x+ → = , which are not equal, 1 ( )lim x f x → does not exist. l. 4 ( )lim x f x−→ As x approaches 1 from the left side, the value of the function approaches 5. Therefore, 4 ( ) 5lim x f x− → = m. 4 ( )lim x f x+→ As x approaches 1 from the right side, the value of the function approaches 5. Therefore 4 ( ) 5lim x f x+→ = n. 4 ( )limx f x → Since 4 ( ) 5lim x f x−→ = and 4 ( ) 5lim x f x+→ = , which are equal, 4 ( ) 5limx f x → = . Example (g), (n) NOTE 1. A limit will exist when two pieces of the graph meet at a point. 2. A limit will not exist when we have: i. A vertical asymptote. Example (h) ii. A jump or step in the graph. Example (j)
Unit 1: Limits and Continuity of a Function 1.13 L i m i t s U s i n g A l g e b r a i c M e t h o d M e t h o d f o r c o m p u t i n g t h e lim ( ) x a f x → . 1. Substitute x a= into ( ).f x If: a. ( )f a is defined then the limit exists and the value ( )f a is the limit. b. Both the numerator and denominator are found to be zero i.e 0 0 then factorise the function and simply and then substitute. c. Only the denominator is equal zero, i.e 0 a then the limit does not exist. 2. If the function ( )f x is a piecewise function and the rule changes at x a= , calculate the left side and right side limits separately. If these are equal the limit exists and that value is the answer. If they are not equal, the limit does not exist. The following examples will show how to use this method with different types of functions. Evaluate 2 1 9 lim 3x x x→ − + Substitute 1x = into the given function. This gives 2 1 9 8 2 1 3 4 − − = = − + Therefore, 2 1 9 lim 2 3x x x→ − = − + Before we look at our next example, try to recall the factorisation techniques that you learned in MAF11 course. They are i. Common factors ii. Difference of two squares: 2 2 ( )( )x y x y x y− = + − iii. Factoring the form: 2 ax bx c+ + Example 1.1.7
Unit 1: Limits and Continuity of a Function 1.14 Evaluate 2 1 1 lim 1x x x→ − − + . Substitute 1x = − into the given function. This gives 2 ( 1) 1 0 1 1 0 − − = − + Since both numerator and denominator are zero when 1x = − , we factorise the numerator and denominator, cancel out any like factors and then substitute again. 2 1 1 lim 1x x x→ − − + 1 ( 1)( 1) lim ( 1)x x x x→ − − + = + Factorise numerator 1 ( 1)( 1) lim ( 1)x x x x→ − − + = + Cancel out common factor 1 lim ( 1) x x → − = − 1 1 2= − − = − Substitute again Evaluate 2 23 2 15 lim 2 6x x x x x→ + − − Substituting 3x = into the expression yields 0 0 which is undefined. 2 23 3 2 15 ( 5)( 3) lim lim 2 6 2 ( 3)x x x x x x x x x x→ → + − + − = − − Factorise numerator and denominator. 3 ( 5)( 3) lim 2 ( 3)x x x x x→ + − = − Cancel out common factor. 3 5 lim 2x x x→ + = Example 1.1.8 Example 1.1.9 N O T E : To find the value of a negative number raised to a power, you MUST USE BRACKETS as shown above.
Unit 1: Limits and Continuity of a Function 1.15 3 5 2(3) + = Substitute again. 8 4 6 3 = = Evaluate 22 2 lim 3 12x x x→ − − Substituting 2x = into the expression yields 0 0 which is undefined. Hence, we need to factorise and cancel the common factor. 2 22 2 2 2 2 2 lim lim lim 3 12 3( 4) 3( 2)( 2)x x x x x x x x x x→ → → − − − = = − − − + 2 1 lim 3( 2)x x→ = + 2 1 1 1 1 lim 3( 2) 3(2 2) 3(4) 12x x→ = = = + + Evaluate 23 3 lim 9x x x→ + − When we substitute 3x = into the expression, only the denominator is zero. 23 3 3 3 6 lim 9 9 9 0x x x→ + + = = − − Therefore, we conclude that the limit does not exist. Example 1.1.11 Example 1.1.10 N O T E : It is good to leave the answer in the fraction form than in the decimals. Fractions are more précised than decimals.
Unit 1: Limits and Continuity of a Function 1.16 ( ) ( )24 2 ( 4) 2 lim 2 x x x x → − + − Evaluate 4 4 lim 2x x x→ − − Substituting 4x = into the expression yields 0 0 which is undefined. For these type of expressions we rationalize the denominator (or in some problems the numerator) by multiplying both the numerator and denominator by conjugate of the denominator ( )2x + . Therefore, we have: 4 24 lim 2 2x xx x x→ +− × − + Multiply. In the denominator we have used the formula. 2 2 ( )( )a b a b a b− + = − . ( ) 4 ( 4) 2 lim 4x x x x→ − + − Cancel out common factor. 4 lim 4 2 2 2 4 x → + = + = Substitute. Evaluate 1 ( )lim x f x → where 2 2 3, 1 ( ) 1 2 , 1 x x f x x x − < = − ≥ . Since ( )f x is a piecewise function, consider left and right limits separately. Example 1.1.12 Example 1.1.13 N O T E : 1. For the left limit we use the equation with < or symbol. 2. Likewise for the right limit we use the equation with > or symbol.
Unit 1: Limits and Continuity of a Function 1.17 The left limit is given by: 1 1 lim ( ) lim (2 3) 2(1) 3 1 x x f x x− − → → = − = − = − . The right limit is given by: 2 2 1 1 lim ( ) lim (1 2 ) 1 2(1) 1 x x f x x+ + → → = − = − = − Since both left and right limits are equal to 1− , it follows that: 1 lim ( ) 1 x f x → = − . A function ( )f x is defined as: 2 1, 2 ( ) 1, 2 x x f x x x − > − = + ≤ − . Evaluate 2 lim ( ). x f x →− The left limit is given by: 2 2 2 2 lim ( ) lim ( 1) ( 2) 1 5 x x f x x− − → − → − = + = − + = . The right limit is given by: 2 2 lim ( ) lim ( 1) 2 1 3 x x f x x+ + →− → − = − = − − = − Hence, 2 lim ( ) x f x → − does not exist. A function ( )f x is defined as: 2 2 1, 1 ( ) 1, 1 1 2 , 1 x x f x x x x x + > − = + = − − < − . Evaluate 1 lim ( ) x f x →− The left limit is given by: 2 2 1 1 lim ( ) lim (1 2 ) 1 2( 1) 1 2 1 x x f x x− − → − → − = − = − − = − = − Example 1.1.14 Example 1.1.15
Unit 1: Limits and Continuity of a Function 1.18 The right limit is given by: 1 1 lim ( ) lim (2 1) 2( 1) 1 2 1 1 x x f x x+ + → − → − = + = − + = − + = − Since the left and right limits are equal, we can write 1 lim ( ) 1. x f x →− = − 1. Complete the table and use the result to estimate the limit numerically. a 2 1 2 3 lim 2 2x x x x→ + − − x 0.9 0.99 0.999 1 1.001 1.01 1.1 ( )f x b. 2 22 5 6 lim 4x x x x→− − + − x 2.1− 2.01− 2.001− 2− 1.999− 1.99− 1.9− ( )f x 2. The graph of a function ( )y f x= is shown below. Exercise 1.1
Unit 1: Limits and Continuity of a Function 1.19 Use the graph to evaluate the following. a. i. ( 3)f − ii. 3 lim ( ) x f x− → − iii. 3 lim ( ) x f x+ → − iv. 3 lim ( ) x f x →− b. i. ( 1)f − ii. 1 lim ( ) x f x− →− iii. 1 lim ( ) x f x+ →− iv. 1 lim ( ) x f x →− c. i. (2)f ii. 2 lim ( ) x f x− → iii. 2 lim ( ) x f x+ → iv. 2 lim ( ) x f x → d. i. (4)f ii. 4 lim ( ) x f x− → iii. 4 lim ( ) x f x+ → iv. 4 lim ( ) x f x → 3. A function ( )f x is defined as: 2 3 , 1 ( ) 1, 1 2 3, 1 x x f x x x x − − > − = = − + < − Find the value of the following. i. ( 1)f − ii. 1 lim ( ) x f x− →− iii. 1 lim ( ) x f x+ →− iv. 1 lim ( ) x f x →− 4. Find the 1 lim ( ) x f x → by evaluating the corresponding one-sided limits for the given function 2 2 4, 1 ( ) . 1 2 , 1 x x f x x x − < = − ≥ 5. Given 2 2 , 1 ( ) , 1, 0 x x f x x x ≤ = + > find each of the following limits. If the limit does not exist, explain why. i. 0 lim ( ) x f x− → ii. 0 lim ( ) x f x+ → iii. 0 lim ( ) x f x →
Unit 1: Limits and Continuity of a Function 1.20 6. Evaluate the following limits algebraically. a. 2 1 lim (3 5) x x → − b. 24 3 lim 1x x x→ − + c. 2 2 4 (2 ) lim 4x x x x→ − − + d. 26 6 lim 36x x x→ − − e. 2 21 3 2 1 lim 3 3x x x x→ − + − − f. 2 22 2 8 lim 3 7 2x x x x→ − − + g. 2 2 6 lim 3 6x x x x→ − − − + h. 2 2 0 2( ) 5( ) (2 5 ) lim h x h x h x x h→ + − + − − i. 9 7 2 lim 3x x x→ + − + j. 4 4 lim 2x x x→ − − k. 9 3 lim 9x x x→ − − Don’t forget to check your answers at the back of this section!
Unit 1: Limits and Continuity of a Function 1.21 1.2 Continuity of a Function The word continuous is defined in the dictionary as ‘going on without a break’. This word has a very similar meaning in mathematics when we talk about functions whose graphs form continuous curves without gaps, breaks or holes. The graph of a function f will have a hole or a break in it at a point c if any of the following situations occur: • The function f is undefined at c (Figure1.2.1 a). • The limit of ( )f x does not exist as x approaches c (Figures 1.2.1b, 2.4.1c). • The value of the function and the value of the limit at c are different (Figure 1.2.1 d). (a) (b) (c) (d) Figure 1.2.1 This suggests the following definitions. This definition makes use of the concept of limits that we studied in the previous section. D e f i n i t i o n If one or more of the conditions fails to hold, then we say that the function ( )f x discontinuous at the point x c= . We will now look at examples where a function ( )f x is discontinuous at a value x c= and then look at the graph of ( )f x to see what we have at that particular x − value. A function ( )f x is said to be continuous at a point c if the following conditions are satisfied: 1. ( )f c is defined. 2. ( )lim x c f x → exists. 3. ( ) ( )lim x c f x f c → = . c c c c
Unit 1: Limits and Continuity of a Function 1.22 Determine whether the function ( )f x is continuous at the value 1x = ? 2 ( ) 1 f x x = − The function is undefined at 1x = and since only the denominator is zero, the limit does not exist. Let us look at the graph of ( )f x to see what we have at 1x = . The graph is shown below. You can see that we have a vertical asymptote at 1x = . There is a break in the graph of the function. We therefore say that a function will be discontinuous at any vertical (or x ) asymptote. Determine whether the function ( )f x is continuous at the value 2x = − ? 2 4 ( ) 2 x f x x − = + The function is undefined at 2x = − and since both the numerator and denominator are zero, we can calculate the limit of ( )f x at 2x = − . This is shown below. 2 2 2 2 4 ( 2)( 2) lim lim lim 2 2 2 4 2 2x x x x x x x x x→ − →− →− − − + = = − = − − = − + + Example 1.2.1 Example 1.2.2
Unit 1: Limits and Continuity of a Function 1.23 Therefore, the function is undefined at 2x = − , but the limit exists. Let us look at the graph of ( )f x to see what we have at 2x = − . The graph is shown below. You can see that we have a hole at 2x = − . There is a break in the graph of the function. Therefore, we say that a function will be discontinuous at any point where there is a hole in the graph. Determine whether the piecewise function ( )f x is continuous at the value 2x = ? 1, 2 ( ) 2 3, 2 x x f x x x + < = − ≥ The function ( )f x is defined at 2x = since we have (2) 2(2) 3 1f = − = . The limit of ( )f x as 2x → does not exist. This is shown below. We have that, 2 lim ( ) 2 1 3 x f x− → = + = and 2 lim ( ) 2(2) 3 1 x f x+ → = − = . Let us look at the graph of ( )f x to see what we have at 2x = . The graph is shown below. You can see that we have a jump at 2x = . There is a break in the graph of the function. We therefore say that a function will be discontinuous at any point where there is a jump in the graph. Example 1.2.3
Unit 1: Limits and Continuity of a Function 1.24 We can therefore say that a function will be discontinuous when there is a: vertical asymptote. hole. jump. at a particular value x c= . The examples that follow will show you how to determine if a function is continuous at a particular value of x . Determine whether the piecewise function ( )f x is continuous at the value 1x = − ? 2 1 , 1 ( ) 1, 1 x x f x x x + < − = + ≥ − To determine if this function is continuous at 1x = − we need to calculate the values of ( 1)f − and 1 ( )lim x f x → − . To find 1 ( )lim x f x → − , we need to find the one sided limits. Therefore, we have: 2 ( 1) ( 1) 1 1 1 2f − = − + = + = , 1 lim ( ) 1 1 0 x f x− → − = + − = and 2 1 lim ( ) ( 1) 1 1 1 2 x f x+ →− = − + = + = . Therefore, 1 lim ( ) x f x → − does not exist since 1 1 lim ( ) lim ( ) x x f x f x− + → − → − ≠ , thus we say that the function ( )f x is discontinuous at 1x = − . N O T E DO NOT DRAW GRAPHS WHEN ASKED TO DETERMINE if a function is continuous at a certain value. Example 1.2.4
Unit 1: Limits and Continuity of a Function 1.25 Determine whether the piecewise function ( )f x is continuous at the value 2x = ? 2 3, 2 ( ) 1, 2 5, 2 x x f x x x x − < = − = − > To determine if this function is continuous at 2x = we need to calculate the value of (2 )f and 2 lim ( ) x f x → . To find 2 lim ( ) x f x → , we need to find the one sided limits. Therefore, we have: (2 ) 1f = − , 2 lim ( ) 2 3 1 x f x− → = − = − and 2 2 lim ( ) (2) 5 1 x f x+ → = − = − . Therefore, the 2 lim ( ) 1 x f x → = − since 2 2 lim ( ) lim ( ) 1 x x f x f x− + → → = = − ,. We can therefore conclude that the function ( )f x is continuous at 2x = , since: 1. (2)f is defined; 2. 2 lim ( ) x f x → exists; 3. 2 (2) lim ( ) x f f x → = Determine whether the function ( )f x is continuous? 2 1 ( ) ( 2)( 3) x f x x x − = − + The function ( )f x shown above is discontinuous at 2x = and 3x = − , since these are the points where the denominator equals zero. The function is undefined at these points. Example 1.2.5 Example 1.2.6
Unit 1: Limits and Continuity of a Function 1.26 Determine the value of k so that the function ( )f x given below is continuous at 1.x = 2 6 2, 1 ( ) , 1 x x f x kx x − ≤ = > We need to find the value of the following and make them all equal. 1. (1)f 2 (1) 6(1) 2 6 2 4f = − = − = 2. 1 lim ( ) x f x− → 2 1 lim ( ) 6(1) 2 4 x f x− → = − = 3. 1 lim ( ) x f x+ → For the function to be continuous the limit has to equal to 4, therefore the right sided limit should also be equal to 4. Thus, 2 1 lim ( ) (1) 4 x f x k+ → = = Therefore, we have: 4k = . 1. Determine the points where each of the following functions is discontinuous (if any). a. 2 ( ) 2 1f x x= − b. 3 ( ) ( 2) f x x x = + c. 2 4 ( ) 2 x f x x − = + d. 2 3 2 ( ) 1 x f x x + = + Exercise 1.2 Example 1.2.7
Unit 1: Limits and Continuity of a Function 1.27 2. Determine if the following functions are continuous at the given value of x . Give reason(s) for your answer. a. 2 2 , 2 ( ) 4 , 2 x x f x x x − > − = − ≤ − , at 2x = − . b. 2 2, 1 ( ) 2, 1 3 2 , 1 x x f x x x x − < − = − = − − − > − , at 1x = − . c. 2 2 3, 1 ( ) 2, 1 1 2 , 1 x x f x x x x x − > = − = − < , at 1x = . d. 2 3 , 1 ( ) 3 , 1 3, 1 x x f x x x x − > − = = − + < − , at 1x = − 3. Determine the value of k so that the function ( )f x given below will be continuous at 1x = − . 2 3 , 1 ( ) 1, 1 2 , 1 kx x f x x l x x − > − = = − + < −
Unit 1: Limits and Continuity of a Function 1.28 4. The graph of a function ( )f x is shown below. For what values of x is ( )f x discontinuous? Don’t forget to check your answers at the back of this section!
Unit 1: Limits and Continuity of a Function 1.29 Summary 1. a. The phrase “ x approaches a from the left” is shown by x a − → . The minus sign means from the negative side of a , which is the left. b. The phrase “ x approaches a from the right” is shown by x a + → . The positive sign means from the positive side of a , which is the right. 2. Therefore, the two-sided limit lim ( ) x a f x L → = exists only if both one-sided limits are the same. That is only if the value of ( )f x approaches the same number as the value of x approaches a from either side. 3. Do not try to solve limit problems by drawing graphs. 4. When two different pieces of graph meet at a point, the limit exists for the given x value and is equal to the y value of the point. 5. A limit will not exist when we have a: a. Vertical asymptote. b. Jump or step in the graph. 6. Method for computing the lim ( ) x a f x → . a. Substitute x a= into ( ).f x If i. ( )f a is defined the limit exists and the value ( )f a is the limit. ii. Both the numerator and denominator are found to be zero i.e 0 0 then factorise the function and simplify and then substitute. iii. Only the denominator is equal zero, i.e 0 a then the limit does not exist. b. If the function ( )f x is a piecewise function and the rule changes at x a= , calculate the left side and right side limits separately. If these are equal the limit exists and that value is the answer. If they are not equal, the limit does not exist.
Unit 1: Limits and Continuity of a Function 1.30 7. A function ( )f x is continuous at the value x c= , if the following three conditions are satisfied: 1. ( )f c is defined. 2. lim ( ) x c f x → exists. 3. lim ( ) ( ) x c f x f c → = . 8. We can therefore say that a function will be discontinuous when there is a: 1. vertical asymptote. 2. hole. 3. jump. at a particular value x c= . 9. Do not draw graphs to determine if a function is continuous at a given x − value. 10. You must use the given c value and not the letter c that is used in the definition when giving the reasons why a function is continuous at a given x − value.
Unit 1: Limits and Continuity of a Function 1.31 Unit 1 Revision Exercise 1. Evaluate the following limits. a. 2 2 5 14 lim 2 4x x x x→ + − − b. 1 1 lim 1x x x→ − − c. 2 2 0 2( ) 5( ) (2 5 ) lim h x h x h x x h→ + + + − + d. 2 23 2 15 lim 2 6x x x x x→ + − − e. 2 23 9 lim 3 8 3x x x x→ − − − 2. Compute the limit if it exists. a. The 1 lim ( ) x f x → − where ( )f x = 2 1 , 1 1, 1 x x x x + < − + ≥ − b. The 2 lim ( ) x f x → where ( )f x = 2 4, 2 2, 2 6, 2 x x x x x − < − = − > 3. Determine if the following functions are continuous at the given value of x . a. 2 2 , 1 ( ) 3 , 1 2, 1 x x f x x x x − > − = = − + < − , at 1x = − . b. 2 2 1, 2 ( ) 5 , 2 1, 2 x x f x x x x − > − = − = − − < − , at 2x = − .
Unit 1: Limits and Continuity of a Function 1.32 4. Determine the value of k so that the function ( )f x given below is continuous at 1x = − . 3 , 1 ( ) 4, 1 1, 1 x x f x x kx x − ≤ − = = − + > − Don’t forget to check your answers at the back of this section.