MATH 12002 - CALCULUS I 1.6: Limits at Infinity Professor Donald L. - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 1.6: Limits at Infinity Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 13

Introduction to Limits at Infinity 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 now consider limits at infinity. A function y = f ( x ) has limit L at infinity if the values of y become arbitrarily close to L when x becomes large enough. Our basic definition is: Definition Let y = f ( x ) be a function and let L be a number. The limit of f as x approaches + ∞ is L if y can be made arbitrarily close to L by taking x large enough (and positive). We write x → + ∞ f ( x ) = L . lim Compare this to the definition of lim x → a f ( x ) = L . The definition means that the graph of f is very close to the horizontal line y = L for large values of x . D.L. White (Kent State University) 2 / 13

Introduction to Limits at Infinity Most of the functions we study that have finite limits at infinity are quotients of functions. To evaluate these limits at infinity, we will use the following idea. Basic Principle If c is a real number and r is any positive rational number, then c lim x r = 0 . x → + ∞ If c is a real number and r is any positive rational number such that x r is defined for x < 0 , then c lim x r = 0 . x →−∞ This is used as in the following examples. D.L. White (Kent State University) 3 / 13

Examples Example Find 2 x 3 + 5 lim 7 x 3 + 4 x + 3 x → + ∞ if the limit exists. Before we solve this problem, notice that 2 x 3 + 5 and 7 x 3 + 4 x + 3 both approach + ∞ as x → + ∞ and so x → + ∞ (2 x 3 + 5) lim 2 x 3 + 5 lim 7 x 3 + 4 x + 3 � = x → + ∞ (7 x 3 + 4 x + 3) . lim x → + ∞ In order to evaluate the limit, we will first use an algebraic manipulation to turn this into an expression whose limit is the quotient of the limits of the numerator and denominator. We will then use the Basic Principle to evaluate these limits. D.L. White (Kent State University) 4 / 13

Examples Solution Divide the numerator and denominator by the highest power of x that appears in the denominator; in this case x 3 . That is, (2 x 3 + 5) 1 2 x 3 + 5 x 3 lim = lim 7 x 3 + 4 x + 3 (7 x 3 + 4 x + 3) 1 x → + ∞ x → + ∞ x 3 2 x 3 x 3 + 5 x 3 = lim 7 x 3 x 3 + 4 x x 3 + 3 x → + ∞ x 3 2 + 5 x 3 = lim . 7 + 4 x 2 + 3 x → + ∞ x 3 [Continued → ] D.L. White (Kent State University) 5 / 13

Examples Solution [continued] 5 4 3 Now the Basic Principle says that x 3 , x 2 , and x 3 all approach 0 as x → ∞ , and so 2 x 3 + 5 2 + 5 x 3 lim = lim 7 x 3 + 4 x + 3 7 + 4 x 2 + 3 x → + ∞ x → + ∞ x 3 5 x → + ∞ 2 + lim lim x 3 x → + ∞ = 4 3 x → + ∞ 7 + lim lim x 2 + lim x 3 x → + ∞ x → + ∞ 7 + 0 + 0 = 2 2 + 0 = 7 . 2 x 3 + 5 7 x 3 + 4 x + 3 = 2 Hence lim 7 . x → + ∞ D.L. White (Kent State University) 6 / 13

Examples Example Find √ x 3 x 4 + 3 lim 5 − 8 x 4 x →−∞ if the limit exists. D.L. White (Kent State University) 7 / 13

Examples Solution √ x = x 1 / 3 is defined for x < 0 and the highest power of x in Observe that 3 the denominator is x 4 . We have √ x √ x ) 1 (3 x 4 + 3 x 4 + 3 3 x 4 lim = lim 5 − 8 x 4 (5 − 8 x 4 ) 1 x →−∞ x →−∞ x 4 3 x 4 x 4 + x 1 / 3 x 4 = lim x 4 − 8 x 4 5 x →−∞ x 4 1 3 + x 11 / 3 = lim 5 x 4 − 8 x →−∞ 0 − 8 = − 3 3 + 0 = 8 . √ x 3 x 4 + 3 = − 3 Hence lim 8 . 5 − 8 x 4 x →−∞ D.L. White (Kent State University) 8 / 13

Examples Example Find 5 x 4 + x 3 lim 2 x 5 − 4 x 2 x → + ∞ if the limit exists. D.L. White (Kent State University) 9 / 13

Examples Solution The highest power of x in the denominator is x 5 . We have 5 x 4 + x 3 (5 x 4 + x 3 ) 1 x 5 lim = lim 2 x 5 − 4 x 2 (2 x 5 − 4 x 2 ) 1 x → + ∞ x → + ∞ x 5 5 x 4 x 5 + x 3 x 5 = lim 2 x 5 x 5 − 4 x 2 x → + ∞ x 5 5 x + 1 x 2 = lim 2 − 4 x → + ∞ x 3 2 − 0 = 0 0 + 0 = 2 = 0 . 5 x 4 + x 3 Hence lim 2 x 5 − 4 x 2 = 0 . x → + ∞ D.L. White (Kent State University) 10 / 13

Examples Example Find 3 x 5 + 2 lim 4 x 2 − 7 x x →−∞ if the limit exists. D.L. White (Kent State University) 11 / 13

Examples Solution The highest power of x in the denominator is x 2 . We have (3 x 5 + 2) 1 3 x 5 + 2 x 2 lim = lim 4 x 2 − 7 x (4 x 2 − 7 x ) 1 x →−∞ x →−∞ x 2 3 x 5 x 2 + 2 x 2 = lim 4 x 2 x 2 − 7 x x →−∞ x 2 3 x 3 + 2 x 2 = lim 4 − 7 x →−∞ x x → 4 − 0 = 4 , while x 3 → −∞ . x 2 → 0 and 4 − 7 2 Now as x → −∞ , 3 x 5 +2 Hence we have an infinite limit at infinity; lim 4 x 2 − 7 x = −∞ . x →−∞ D.L. White (Kent State University) 12 / 13

General Conclusion We can generalize these methods to obtain the following result. Theorem Let f ( x ) = P ( x ) Q ( x ) be a rational function (P ( x ) , Q ( x ) are polynomials). If the degrees of P ( x ) and Q ( x ) are equal , then x →−∞ f ( x ) = p x → + ∞ f ( x ) = lim lim q , where p is the coefficient of the highest degree term of P ( x ) and q is the coefficient of the highest degree term of Q ( x ) . If the degree of P ( x ) is less than the degree of Q ( x ) , then x → + ∞ f ( x ) = 0 and lim x →−∞ f ( x ) = 0 . lim If the degree of P ( x ) is greater than the degree of Q ( x ) , then x → + ∞ f ( x ) and lim x →−∞ f ( x ) are infinite. lim D.L. White (Kent State University) 13 / 13