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MATH 12002 - CALCULUS I §1.6: Limits at Infinity

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MATH 12002 - CALCULUS I §1.6: Limits at Infinity MATH 12002 - CALCULUS I x1.6: 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 f (x) = L required a and L to be real numbers. x!a 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 +1 is L if y can be made arbitrarily close to L by taking x large enough (and positive). We write lim f (x) = L. x!+1 Compare this to the definition of lim f (x) = L. x!a 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 = 0: x!+1 xr If c is a real number and r is any positive rational number such that xr is defined for x < 0, then c lim = 0: x→−∞ xr This is used as in the following examples. D.L. White (Kent State University) 3 / 13 Examples Example Find 2x3 + 5 lim x!+1 7x3 + 4x + 3 if the limit exists. Before we solve this problem, notice that 2x3 + 5 and 7x3 + 4x + 3 both approach +1 as x ! +1 and so 3 2x3 + 5 lim (2x + 5) lim 6= x!+1 : x!+1 7x3 + 4x + 3 lim (7x3 + 4x + 3) x!+1 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 x3. That is, 3 3 1 2x + 5 (2x + 5) 3 lim = lim x x!+1 7x3 + 4x + 3 x!+1 3 1 (7x + 4x + 3) x3 3 2x + 5 = lim x3 x3 x!+1 7x3 4x 3 x3 + x3 + x3 2 + 5 = lim x3 : x!+1 4 3 7 + x2 + x3 [Continued !] D.L. White (Kent State University) 5 / 13 Examples Solution [continued] 5 4 3 Now the Basic Principle says that x3 , x2 , and x3 all approach 0 as x ! 1, and so 3 5 2x + 5 2 + 3 lim = lim x x!+1 7x3 + 4x + 3 x!+1 4 3 7 + x2 + x3 5 lim 2 + lim 3 = x!+1 x!+1 x 4 3 lim 7 + lim + lim x!+1 x!+1 x2 x!+1 x3 2 + 0 2 = = : 7 + 0 + 0 7 2x3 + 5 2 Hence lim = . x!+1 7x3 + 4x + 3 7 D.L. White (Kent State University) 6 / 13 Examples Example Find p 3x4 + 3 x lim x→−∞ 5 − 8x4 if the limit exists. D.L. White (Kent State University) 7 / 13 Examples Solution p Observe that 3 x = x1=3 is defined for x < 0 and the highest power of x in the denominator is x4. We have p p 4 3 4 3 1 3x + x (3x + x) 4 lim = lim x x→−∞ 5 − 8x4 x→−∞ 4 1 (5 − 8x ) x4 4 1=3 3x + x = lim x4 x4 x→−∞ 5 8x4 x4 − x4 3 + 1 = lim x11=3 x→−∞ 5 x4 − 8 3 + 0 3 = = − : 0 − 8 8 p 3x4+ 3 x 3 Hence lim 4 = − . x→−∞ 5−8x 8 D.L. White (Kent State University) 8 / 13 Examples Example Find 5x4 + x3 lim x!+1 2x5 − 4x2 if the limit exists. D.L. White (Kent State University) 9 / 13 Examples Solution The highest power of x in the denominator is x5. We have 4 3 4 3 1 5x + x (5x + x ) 5 lim = lim x x!+1 2x5 − 4x2 x!+1 5 2 1 (2x − 4x ) x5 4 3 5x + x = lim x5 x5 x!+1 2x5 4x2 x5 − x5 5 + 1 = lim x x2 x!+1 4 2 − x3 0 + 0 0 = = = 0: 2 − 0 2 5x4+x3 Hence lim 5 2 = 0. x!+1 2x −4x D.L. White (Kent State University) 10 / 13 Examples Example Find 3x5 + 2 lim x→−∞ 4x2 − 7x if the limit exists. D.L. White (Kent State University) 11 / 13 Examples Solution The highest power of x in the denominator is x2. We have 5 5 1 3x + 2 (3x + 2) 2 lim = lim x x→−∞ 4x2 − 7x x→−∞ 2 1 (4x − 7x) x2 5 3x + 2 = lim x2 x2 x→−∞ 4x2 7x x2 − x2 3x3 + 2 = lim x2 x→−∞ 7 4 − x 2 7 3 Now as x ! −∞, x2 ! 0 and 4 − x ! 4 − 0 = 4, while x ! −∞. 3x5+2 Hence we have an infinite limit at infinity; lim 2 = −∞. x→−∞ 4x −7x D.L. White (Kent State University) 12 / 13 General Conclusion We can generalize these methods to obtain the following result. Theorem P(x) Let f (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 p lim f (x) = lim f (x) = ; x!+1 x→−∞ 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 lim f (x) = 0 and lim f (x) = 0. x!+1 x→−∞ If the degree of P(x) is greater than the degree of Q(x), then lim f (x) and lim f (x) are infinite. x!+1 x→−∞ D.L. White (Kent State University) 13 / 13.
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