Section 2.6 Limits at infinity and infinite limits 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

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Section 2.6 Limits at infinity and infinite its 2 Lectures College of Science MATHS 0: Calculus I (University of Bahrain) Infinite Limits / 29

Finite its as ±. 2 Horizontal Asympotes. 3 Infinite its. 4 Vertical Asymptotes. (University of Bahrain) Infinite Limits 2 / 29

Motivation Eample Consider the function f () =. The graph of the function is y Question: What happen if is sufficiently large number (i.e., approaches )? In other words, what is? From the graph we can easily see that = 0 and = 0 (University of Bahrain) Infinite Limits 3 / 29

Continue... Arithmetic at infinity: + =. 2 k = (k > 0). 3 k = (k < 0). 4 5 ± = 0. =?. (Calculus ). 6 =?, 0 =?, =?. (Calculus 2) (University of Bahrain) Infinite Limits 4 / 29

ing the it of a rational function f () To find the it ±, we have g() Substitute directly by = ± in f (). If you get a real number or g() ±, then that is the it. 2 If you get undefined values such as 0 0 or, we take the highest power of in the numerator and the highest power of in the denominator as common factor and we proceed. (University of Bahrain) Infinite Limits 5 / 29

Eample Solution: Direct substitution gives 3 2 2 5 2 + 4 + 3( ) 2 ( ) 2 5( ) 2 + 4( ) + 3 2 2 5 2 + 4 + = = 2 undefined! 3 2 2 5 2 + 4 + ( 3 2 ) 2 ( 2 5 + 4 + = = ( 3 ) 2 2 ) 2 ) 2 ( 5 + 4 + (3 0 0) (5 + 0 + 0) = 3 5 (University of Bahrain) Infinite Limits 6 / 29

Eercise 2 5 2 2 + 9 2 (University of Bahrain) Infinite Limits 7 / 29

Eample 3 3 + 7 2 2 Solution: Direct substitution gives 3 + 7 ( ) 2 2 = undefined! 3 + 7 2 2 = 3 + 7 2 2 ( ) 3 + 7 = ( 2 2 ( ) 3 + 7 = ( 2 2 ) = 2 ) (3 + 0) ( 0) = 0 (University of Bahrain) Infinite Limits 8 / 29

Eample 4 3 8 2 2 + Solution: Direct substitution gives ( ) 3 8 2( ) 2 + = undefined! 3 8 2 2 + = 3 8 2 2 + 3 ( ) 8 = 3 ( 2 2 + ) ( 2 ) 8 = ( 3 2 + = 2 ) (University of Bahrain) Infinite Limits 9 / 29

Eercise 5 5 3 2 + 9 8 + 8 (University of Bahrain) Infinite Limits 0 / 29

Eample 6 3 2 + 3 5 Solution: Direct substitution gives 3( )2 + = 3( ) 5 undefined! 3 2 + 3 2 + = 3 5 3 5 ( 2 3 + ) = 2 ( 3 5 ) = (3 ) + = 2 ( 3 5 ) = (3 ) + = ( 2 3 5 ) = (3 2 + ) ( 3 5 (3 + ( 3 5 ) (University of Bahrain) Infinite Limits / 29 2 ) 2 )

Eercise 7 3 2 + 3 5 Solution: Direct substitution gives 3( )2 + = 3( ) 5 3 2 + 3 2 + = 3 5 3 5 ( 2 3 + = ( 3 5 2 ) undefined! ) = 2 ) (3 2 + ) ( 3 5 (3 + (3 + = ( 3 5 ) = ( 3 5 ) (3 = ) + ( 2 5 ) = 3 (University of Bahrain) Infinite Limits 2 / 29 2 ) 2 )

Multiplying by the conjugate Eample 8 ( 2 + ) Solution: Direct substitution gives ( ) ( ) 2 + = undefined! ( 2 + ) = ( 2 + ( ) 2 + + ) ( 2 + + ) = 2 + 2 ( = 2 + + ) ( 2 + + ) = 0 (University of Bahrain) Infinite Limits 3 / 29

Eercise 9 ( ) 2 + 6 (University of Bahrain) Infinite Limits 4 / 29

2 - Horizontal Asymptotes Motivational Eample: Consider the function f () = 2. Then we have 2 + f () = and f () = In this case, the line y = is called a horizontal asymptote. y (University of Bahrain) Infinite Limits 5 / 29

Definition 0 The line y = L is called a horizontal asymptote of the curve y = f () if either f () = L and f () = L Eample the horizontal asymptote of the function 9 f () = 4 2 + 3 + 2 we need to find both f () and f () (University of Bahrain) Infinite Limits 6 / 29

9 4 2 + 3 + 2 = Hence y = 2 = = = 9 3 2 + 3 + ( ) 9 ( 2 4 + 3 + 2 ) = ( ) 2 9 (4 + 3 + 2 ) = ( ) 2 9 (4 + 3 + 2 ) = 2 2 ( 9 ) (4 2 + 3 ( 9 ) (4 + 3 + 2 ) 2 + 2 ) 2 is a horizontal asymptote. Now we compute f () to get f () = 2 asymptote. and so we have y = 2 is also a horizontal (University of Bahrain) Infinite Limits 7 / 29

Motivation Eample Consider the function f () =. The graph of the function is y Question: What is 0 + and 0? From the graph we can easily see that 0 + = ( ) 0 + = and 0 = ( ) 0 + = (University of Bahrain) Infinite Limits 8 / 29

Eample 2 Solution: Direct substitution gives 3 0 3 + undefined! So we need to find whether it is 0 + or 0. 3 + = 3 0 + = (University of Bahrain) Infinite Limits 9 / 29

Eercise 3 Solution: Direct substitution gives 3 0 3 undefined! So we need to find whether it is 0 + or 0. 3 = 3 0 = (University of Bahrain) Infinite Limits 20 / 29

Eample 4 + Solution: Direct substitution gives 2 0 2 + undefined! So we need to find whether it is 0 + or 0. + 2 + = 2 0 + = 2 = (University of Bahrain) Infinite Limits 2 / 29

Eercise 5 Solution: Direct substitution gives 3 0 3 2 + 2 undefined! So we need to find whether it is 0 + or 0. 3 2 + 2 = 3 0 = (University of Bahrain) Infinite Limits 22 / 29

Eample 6 4 Solution: Direct substitution gives 8 0 2 2 6 undefined! So we need to find whether it is 0 + or 0. 4 2 2 6 = 8 0 = (University of Bahrain) Infinite Limits 23 / 29

Eample 7 2 2 + 2 4 + 4 Solution: Direct substitution gives 0 0 undefined! So we need to factor first using the methods of Section 2.2. 2 2 + 2 4 = ( 2) 2 + ( 2)( 2) = So we need to find whether it is 0 + or 0. 2 + 2 = 0 + = 2 2 + (University of Bahrain) Infinite Limits 24 / 29

Eercise 8 3 3 2 9 Solution: Direct substitution gives 9 undefined! 0 So we need to find whether it is 0 + or 0 and for that we find the right and the left its. 3 3 + 2 9 = 9 0 + = Since 3 + 3 3 2 9 = 3 3 3 2 9 = 9 0 = 3, we have 2 9 3 2 9 Does Not Eist (University of Bahrain) Infinite Limits 25 / 29

Eample 9 2 ( ) 2 Solution: Direct substitution gives 0 0 undefined! So we need to factor first using the methods of Section 2.2. 2 ( ) 2 = ( )( + ) ( )( ) = + So we need to find whether it is 0 + or 0 and for that we find the right and the left its. + + = 2 0 + = + = 2 0 = Since + 2 ( ) 2 = 2 ( ) 2, we have 2 ( ) 2 Does Not Eist (University of Bahrain) Infinite Limits 26 / 29

4 - Vertical Asymptotes Motivational Eample: Consider the function f () = +3 2. Then we have f () = and f () = 2 + 2 In this case, the line = 2 is called a vertical asymptote. y (University of Bahrain) Infinite Limits 27 / 29

Definition 20 The line = a is called a vertical asymptote of the curve y = f () if either f () = ± or f () = ± + a a To find the vertical asymptote for a rational function, we need to cancel any common factor first and we find where the denominator is zero. Eample 2 the vertical asymptote of the function f () = 8 2 4 2 4 = 0 = 2, = 2. Since none of these is a zero for the numerator, then both are vertical asymptote. (University of Bahrain) Infinite Limits 28 / 29

Eercise 22 the vertical asymptote of the function 2 4+3 ( )( 3) 2 ( )(+) ( 3) (+) asymptote. f () = 2 4 + 3 2 =. So only = is a vertical (University of Bahrain) Infinite Limits 29 / 29