MATH 12002 - CALCULUS I § 1.6: Infinite Limits Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 9
Introduction to Infinite Limits Our definition of lim x → a f ( x ) = L required a and L to be real numbers. In this section, we expand the definition to allow a to be infinite (limits at infinity) or L to be infinite (infinite limits). We first consider infinite limits. A function y = f ( x ) has an infinite limit at x = a if the values of y become arbitrarily large (either positive or negative) when x is close to a . Our basic definition is: Definition Let y = f ( x ) be a function and let a be a number. The limit of f as x approaches a is + ∞ if y can be made arbitrarily large (and positive) by taking x close enough to a , but not equal to a . We write lim x → a f ( x ) = + ∞ . Compare this to the definition of lim x → a f ( x ) = L . D.L. White (Kent State University) 2 / 9
Introduction to Infinite Limits Most of the functions we study that have infinite limits at a point x = a are quotients of functions. To evaluate these infinite limits, we will use the following idea. Basic Principle Let f ( x ) = N ( x ) / D ( x ) . If lim x → a D ( x ) = 0 and lim x → a N ( x ) � = 0 , then the one-sided limits at x = a of f ( x ) are infinite; that is N ( x ) N ( x ) lim D ( x ) = ±∞ and lim D ( x ) = ±∞ . x → a + x → a − N ( x ) If the signs are the same, then lim D ( x ) = ±∞ . x → a N ( x ) If the signs are different, then lim D ( x ) does not exist. x → a D.L. White (Kent State University) 3 / 9
Introduction to Infinite Limits The Basic Principle says that in order to find lim x → a f ( x ), where f ( x ) = N ( x ) D ( x ) with lim x → a D ( x ) = 0 and lim x → a N ( x ) � = 0, we only need to determine the signs of lim x → a − f ( x ) and lim x → a + f ( x ). That is, 1 Determine the sign of y = f ( x ) = N ( x ) D ( x ) as x → a − . If +, then lim x → a − f ( x ) = + ∞ . If − , then lim x → a − f ( x ) = −∞ . 2 Determine the sign of y = f ( x ) = N ( x ) D ( x ) as x → a + . If +, then lim x → a + f ( x ) = + ∞ . If − , then lim x → a + f ( x ) = −∞ . 3 If lim x → a − f ( x ) = lim x → a + f ( x ) = ±∞ , then lim x → a f ( x ) = ±∞ . If lim x → a − f ( x ) � = lim x → a + f ( x ), then lim x → a f ( x ) does not exist. D.L. White (Kent State University) 4 / 9
Examples Example Find x 2 + 7 lim x 2 − 8 x + 15 x → 3 if the limit exists or is ±∞ . Solution First notice that as x → 3 , x 2 + 7 → 3 2 + 7 = 16 � = 0 and x 2 − 8 x + 15 → 3 2 − 8(3) + 15 = 9 − 24 + 15 = 0 , so the one-sided limits are infinite. [Continued → ] D.L. White (Kent State University) 5 / 9
Examples Solution [continued] Notice also that x 2 − 8 x + 15 = ( x − 3)( x − 5) , so f ( x ) = x 2 +7 ( x − 3)( x − 5) . x → 3 − f ( x ) : As x → 3 − lim 1 x 2 + 7 → 16 x − 5 → − 2 x − 3 → 0 − , x 2 +7 and so f ( x ) = ( x − 3)( x − 5) is positive. Hence x → 3 − f ( x ) = + ∞ . lim [Continued → ] D.L. White (Kent State University) 6 / 9
Examples Solution [continued] x → 3 + f ( x ) : As x → 3 + lim 2 x 2 + 7 16 → x − 5 → − 2 0 + , x − 3 → x 2 +7 and so f ( x ) = ( x − 3)( x − 5) is negative. Hence x → 3 + f ( x ) = −∞ . lim 3 Since lim x → 3 − f ( x ) � = lim x → 3 + f ( x ) , x → 3 f ( x ) DOES NOT EXIST. lim D.L. White (Kent State University) 7 / 9
Examples Example Find x 3 − 9 lim x 2 − 4 x + 4 x → 2 if the limit exists or is ±∞ . Solution First notice that as x → 2 , x 3 − 9 → 2 3 − 9 = − 1 � = 0 and x 2 − 4 x + 4 → 2 2 − 4(2) + 4 = 4 − 8 + 4 = 0 , so the one-sided limits are infinite. [Continued → ] D.L. White (Kent State University) 8 / 9
Examples Solution [continued] Notice also that x 2 − 4 x + 4 = ( x − 2) 2 , so f ( x ) = x 3 − 9 ( x − 2) 2 . In contrast to the previous example, as x → 2 , we have ( x − 2) 2 → 0 , but ( x − 2) 2 is positive whether x > 2 or x < 2 . Therefore, as x → 2 (from either side), x 3 − 9 → − 1 ( x − 2) 2 0 + , → x 3 − 9 and so f ( x ) = ( x − 2) 2 is negative. Hence x → 2 f ( x ) = −∞ . lim D.L. White (Kent State University) 9 / 9
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