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1101 Calculus I Lecture 2.6: Horizontal Asymptotes

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1101 Calculus I Lecture 2.6: Horizontal Asymptotes Calculus Lecture 2.6: Horizontal Asymptotes Page 1 of 4 1101 Calculus I Lecture 2.6: Horizontal Asymptotes Note: We will not be using the precise definition of limits in this course. Vertical asymptotes are when we let x approach a number and the function f becomes arbitrarily large (positive or negative). For example, lim ln x = −∞ means y = ln x has a vertical asymptote at x = 0. x!0+ Horizontal asymptotes are when we let x become arbitrarily large (positive or negative) and the function f approaches a number. This is the more formal way to treat end behaviour, and allows us to examine more complicated functions than we have in the past (before we focussed mostly on rational functions, and said things like \4x5 − 3x + 1 is dominated by 4x5 when x is large"). Definition Let f be a function defined on some interval (a; 1). Then lim f(x) = L means that the values of f(x) can x!1 be made arbitrarily close to L by taking x sufficiently large. Definition Let f be a function defined on some interval (−∞; a). Then lim f(x) = L means that the values of f(x) can x→−∞ be made arbitrarily close to L by taking x sufficiently large negative. Definition The line y = L is called a horizontal asymptote of the curve y = f(x) if either lim f(x) = L; or lim f(x) = L: x!1 x→−∞ For example, the sketch of f(x) = tan−1 x shows two horizontal asymptotes: π π From the sketch, we see lim arctan x = , and lim arctan x = − . x!1 2 x→−∞ 2 Therefore, the horizontal asymptotes of f(x) = arctan(x) are y = π=2, and y = −π=2. Calculus Lecture 2.6: Horizontal Asymptotes Page 2 of 4 Theorems 1 If r > 0 is a rational number, then lim = 0. x!1 xr 1 If r > 0 is a rational number such that xr is defined for all x, then lim = 0. x→−∞ xr Technique: To evaluate a limit at infinity of a rational function or some algebraic functions, where both numerator and denominator are becoming large, we divide the numerator and denominator by the highest power of x in the denominator so we can use the two results above. We don't have to worry about division by zero since we are looking at very large x. 1 Note that is yet another indeterminant form. 1 x2 + 1 Example Evaluate lim . x!1 2x2 − 3x − 2 x2 + 1 1 lim ! indeterminant form, divide by largest power of x in denominator x!1 2x2 − 3x − 2 1 2 1 x + 1 2 = lim · x x!1 2 1 2x − 3x − 2 x2 1 + 1 = lim x2 x!1 1 1 2 − 3 x − 2 x2 1 + 0 = 2 − 3(0) − 2(0) 1 = 2 p Example Evaluate lim ( 9x2 + x − 3x). x!1 p lim ( 9x2 + x − 3x) −! 1 − 1 indeterminant form; need to do some algebra x!1 p p ( 9x2 + x + 3x) = lim ( 9x2 + x − 3x) · p x!1 ( 9x2 + x + 3x) (¨9x¨2 + x −¨9x¨2) = lim p x!1 ( 9x2 + x + 3x) x 1 = lim p ! divide by largest power of x in denominator x!1 ( 9x2 + x + 3x) 1 x 1=x = lim p · x!1 ( 9x2 + x + 3x) 1=x 1 p = lim p since x = x2 when x > 0 x!1 9x2 + x( p1 ) + 3 x2 1 = lim x!1 q 1 9 + x + 3 1 = p 9 + 0 + 3 1 = 6 Calculus Lecture 2.6: Horizontal Asymptotes Page 3 of 4 p 2x2 + 1 Example Find the horizontal asymptotes of the function: f(x) = . 3x − 5 p 2x2 + 1 1 lim = −! direct substitution does not work; indeterminant form x!1 3x − 5 1 p 2x2 + 1 1=x = lim · x!1 3x − 5 1=x p p 2x2 + 1 1= x2 p = lim · since x = x2 for x > 0 x!1 3x − 5 1=x p2 + 1=x2 = lim x!1 3 − 5=x p 2 + 0 = 3 − 0 p 2 = 3 p So y = 2=3 is a horizontal asymptote. p 2x2 + 1 1 lim = −! direct substitution does not work; indeterminant form x→−∞ 3x − 5 −∞ p 2x2 + 1 1=x = lim · x→−∞ 3x − 5 1=x p 2x2 + 1 1=(−px2) p = lim · since x = − x2 for x < 0 x→−∞ 3x − 5 1=x p2 + 1=x2 = − lim x→−∞ 3 − 5=x p 2 + 0 = − 3 − 0 p 2 = − 3 p So y = − 2=3 is a horizontal asymptote. Recall: lim ex = 1 and lim ex = 0. x!1 x→−∞ 1 − ex Example Evaluate lim . x!1 1 + 2ex 1 − ex −∞ lim ! indeterminant form; divide everything by ex, which is same as multiplying by e−x x!1 1 + 2ex 1 1 − ex e−x = lim · x!1 1 + 2ex e−x e−x − 1 = lim x!1 e−x + 2 0 − 1 −1 = = 0 + 2 2 1 − ex 1 − 0 Note that lim = = 1. x→−∞ 1 + 2ex 1 + 2(0) Calculus Lecture 2.6: Horizontal Asymptotes Page 4 of 4 p 9x6 − x Example Evaluate lim . x!1 x3 + 1 p 9x6 − x 1 lim ! indeterminant form x!1 x3 + 1 1 p 6 1 9x − x x3 = lim 3 · 1 x!1 x + 1 3 p x 9x6 − x p1 p = lim x6 since for x > 0, x3 = x6 x!1 1 1 + x3 q 1 9 − x5 = lim 1 x!1 1 + 3 p x 9 − 0 = 1 + 0 = 3 p 9x6 − x Example Evaluate lim . x→−∞ x3 + 1 Note that in this problem, we need lim (9x6 − x) = lim x(9x5 − 1) = (−∞)(−∞) = 1. x→−∞ x→−∞ p 9x6 − x 1 lim ! indeterminant form x→−∞ x3 + 1 −∞ p 6 1 9x − x x3 = lim 3 · 1 x→−∞ x + 1 3 p x 9x6 − x p1 p = − lim x6 since for x < 0, x3 = − x6 x→−∞ 1 1 + x3 q 1 9 − x5 = − lim 1 x→−∞ 1 + 3 p x 9 − 0 = − 1 + 0 = −3 −4x7 + 5 Example Evaluate lim . x!1 (x2 − 2)(2x2 − 1) −4x7 + 5 −∞ lim ! indeterminant form x!1 (x2 − 2)(2x2 − 1) 1 7 1 −4x + 5 4 = lim · x x!1 4 2 1 2x − 5x + 2 x4 −4x3 + 5 1 = lim x4 x!1 1 1 2 − 5 x2 + 2 x4 −4(1) + 5(0) = 2 − 5(0) + 2(0) = −∞ The limit does not exist..
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