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12. Limit at Infinity

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12. Limit at Infinity 12. Limit at infinity Limit at infinity Limit at infinity by inspection 12.1. Limit at infinity by inspection Limit at infinity of polynomial Limit at infinity of quotient We write limx!1 f(x) to represent the height that the graph of f approaches (if any) as x gets ever larger. Here are the possibilities: lim f(x) = L Table of Contents x!1 JJ II J I Page1 of 11 Back lim f(x) = 1 x!1 Print Version Home Page Limit at infinity Limit at infinity by inspection Limit at infinity of polynomial Limit at infinity of quotient lim f(x) does not exist x!1 (If the second graph had turned downward instead, we would have written limx!1 f(x) = −∞.) Table of Contents Similarly, limx→−∞ f(x) denotes the height that the graph of f approaches (if any) as x gets ever smaller. JJ II 12.1.1 Example Use inspection to find the following limits: J I 5 (a) lim x!1 x Page2 of 11 (b) lim 3x5 x→−∞ Back (c) lim cos(2x + 3) x!1 Print Version (d) lim cos(1=x) x!1 Home Page (e) lim tan−1 x x→−∞ Solution Limit at infinity 5 5 Limit at infinity by inspection (a) lim = 0. (We think .) x!1 x large pos. Limit at infinity of polynomial Limit at infinity of quotient (b) lim 3x5 = −∞. (We think 3(large neg.)5 = large neg..) x→−∞ (c) lim cos(2x+3) does not exist. (The cosine of ever larger numbers oscillates between x!1 −1 and 1 and never settles down to any particular value.) (d) lim cos(1=x) = 1. (Since 1=x goes to 0, the expression goes to cos 0 = 1.) x!1 π (e) lim tan−1 x = − . (The graph of tan−1 x is obtained by reflecting across the 45◦ x→−∞ 2 Table of Contents line x = θ the graph of the portion of the graph of tan θ corresponding to the reduced domain (−π=2; π=2)): JJ II J I Page3 of 11 Back Print Version Home Page 12.2. Limit at infinity of polynomial Limit at infinity Limit at infinity by inspection We cannot use inspection right away to decide the limit Limit at infinity of polynomial lim −2x5 + x4 + x2 − 3 : Limit at infinity of quotient x!1 The first term of the polynomial is becoming large and negative trying to make the poly- nomial go to −∞, while the second term is becoming large and positive trying to make the polynomial go to 1. It is like a struggle is going on and it is not clear which term will win (the presence of the other two terms does not make things any better). To find this limit we factor out the highest power of x, namely x5, use a limit law to break the limit up into a product of limits, and then finally use inspection. Table of Contents 12.2.1 Example Evaluate lim −2x5 + x4 + x2 − 3. x!1 JJ II Solution The first step is to factor out the highest power of x, namely x5: J I 1 1 3 lim −2x5 + x4 + x2 − 3 = lim x5 −2 + + − x!1 x!1 x x3 x5 Page4 of 11 1 1 3 = lim x5 · lim −2 + + − ((large pos.) · (about −2)) x!1 x!1 x x3 x5 Back = −∞: Print Version 3 12.2.2 Example Evaluate lim 6 − x − 4x . Home Page x→−∞ Solution We factor out the highest power of x, which is x3: Limit at infinity Limit at infinity by inspection 3 3 6 1 lim 6 − x − 4x = lim x − − 4 Limit at infinity of polynomial x→−∞ x→−∞ x3 x2 6 1 Limit at infinity of quotient = lim x3 · lim − − 4 ((large neg.) · (about −4)) x→−∞ x→−∞ x3 x2 = 1: Close inspection of these two examples reveals a general principle: Table of Contents In finding the limit at infinity of a polynomial, all terms other than the one with the highest power of x can be ignored. JJ II For instance, J I lim −2x5 + x4 + x2 − 3 = lim −2x5 = −∞; x!1 x!1 Page5 of 11 and lim 6 − x − 4x3 = lim −4x3 = 1: x→−∞ x→−∞ Back This technique is not valid unless x is going to either 1 or −∞. Print Version Home Page 12.3. Limit at infinity of quotient Limit at infinity Limit at infinity by inspection In the limit 1 + x + 3x2 Limit at infinity of polynomial lim Limit at infinity of quotient x!1 4x2 − 5 1 both numerator and denominator go to infinity, so the limit has the form 1 . It was stated in 8.3 that inspection alone cannot be used to find limits with this form. The numerator getting large is trying to make the fraction large, while the denominator getting large is trying to make the fraction small, so it is like a struggle is going on and it is not clear whether the numerator or the denominator will win (or some compromise will be reached). To find the limit, we divide both numerator and denominator by the highest power of x that appears in the denominator, namely x2. Table of Contents 1 + x + 3x2 12.3.1 Example Evaluate lim x!1 4x2 − 5 JJ II Solution We divide both numerator and denominator by the highest power of x that J I appears in the denominator, namely x2: Page6 of 11 Back Print Version Home Page Limit at infinity 1 + x + 3x2 Limit at infinity by inspection 2 1 + x + 3x 2 Limit at infinity of polynomial lim = lim x x!1 4x2 − 5 x!1 4x2 − 5 Limit at infinity of quotient x2 1 x 3x2 2 + 2 + 2 = lim x x x x!1 4x2 5 − x2 x2 1 1 2 + + 3 = lim x x x!1 5 4 − Table of Contents x2 0 + 0 + 3 = 4 − 0 JJ II 3 = : 4 J I Page7 of 11 To evaluate this limit, we could also use that fact that, for limits at infinity of a polynomial, one can ignore all terms other than the one with the highest power of x: Back 1 + x + 3x2 3x2 3 3 lim = lim = lim = : Print Version x!1 4x2 − 5 x!1 4x2 x!1 4 4 x3 − x + 7 Home Page 12.3.2 Example Evaluate lim p . x!1 9x6 + 1 Solution As in the previous example, both numerator and denominator are going to in- Limit at infinity finity, so we cannot use inspection alone. We divide numerator and denominator by the Limit at infinity by inspection highest power of x appearing in the denominator. This highest power appears to be x6, Limit at infinity of polynomial but, due to the square root sign, the true highest power appearing in the denominator is x3, so this is what we divide by: Limit at infinity of quotient x3 − x + 7 3 x − x + 7 3 lim p = lim p x x!1 9x6 + 1 x!1 9x6 + 1 x3 x3 x 7 − + x3 x3 x3 = lim r x!1 9x6 + 1 Table of Contents x6 1 7 JJ II 1 − 2 + 3 = lim x x x!1 r 1 9 + J I x6 1 − 0 + 0 = p 9 + 0 Page8 of 11 1 = 3 Back Print Version To evaluate this limit, we could also use that fact that, for limits at infinity of a polynomial, Home Page one can ignore all terms other than the one with the highest power of x: Limit at infinity x3 − x + 7 x3 x3 1 1 Limit at infinity by inspection lim p = lim p = lim = lim = : Limit at infinity of polynomial x!1 9x6 + 1 x!1 9x6 x!1 3x3 x!1 3 3 Limit at infinity of quotient Table of Contents JJ II J I Page9 of 11 Back Print Version Home Page 12 { Exercises Limit at infinity Limit at infinity by inspection Limit at infinity of polynomial Limit at infinity of quotient 12 { 1 Use inspection to find the following limits: −3 (a) lim x!1 x2 − 5 (b) lim 4 − x−3 x→−∞ cos x (c) lim x!1 x Table of Contents (d) lim tan−1(πx − 1) x!1 (e) lim e−x JJ II x→−∞ J I 12 { 2 Find the following limit by factoring out the highest power of x that appears: Page 10 of 11 lim 3 − 2x2 + x6 − 4x7 : x!1 Back 12 { 3 Evaluate the following limit by dividing numerator and denominator by the highest power Print Version of x that appears in the denominator: 8x3 − x + 2 Home Page lim : x!1 7 − x2 + 2x3 Limit at infinity Limit at infinity by inspection 12 { 4 Evaluate the following limit by dividing numerator and denominator by the true highest Limit at infinity of polynomial power of x that appears in the denominator: Limit at infinity of quotient x2 − 3 lim p x!1 3 −8x6 + 7x4 − x . 12 { 5 Find the following limits by ignoring all terms in each polynomial other than the one with the highest power of x. (Note: This works only for limits at infinity.) Table of Contents (a) lim 6x8 − 7x3 + 4x − 9 x→−∞ JJ II 5x − 6x3 (b) lim 3 2 x→−∞ 2x − 8x + 1 J I p 3 27x6 + 5x4 − 3 (c) lim x!1 x2 − 9 Page 11 of 11 Back Print Version Home Page.
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