# 1.5.2 - 1.5 The Limit of a Function part 2 Dr Chenying Wang...

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This preview shows page 1 out of 24 pages. Unformatted text preview: 1.5 The Limit of a Function, part 2 Dr. Chenying Wang Lecturer Department of Mathematics Outline Definition of infinite limits How to find infinite limits. How to find vertical asymptotes. 2 Only for math 140, sections 1 & 020, fall 15 Example 1 Evaluate each of the following limits. 1 lim x!0+ x 1 lim x!0¡ x 1 lim x!0 x x 1/x x 1/x -0.1 -10 0.1 10 -0.01 -100 0.01 100 -0.001 -1000 0.001 1000 -.00001 -10000 0.0001 10000 Solution: We choose values of x that get closer and closer to x=0 from both the left and the right and plug these values into the function. 3 Only for math 140, sections 1 & 020, fall 15 Example 1 - Solution From this table we can see that as we make x smaller and smaller the function gets larger and larger and will retain the same sign that x originally had. We can make the function as large and positive as we want for all x’s sufficiently close to zero while staying positive (i.e. on the right). Likewise, we can make the function as large and negative as we want for all x’s sufficiently close to zero while staying negative (i.e. on the left). To indicate the kind of behavior, we use the following notations. 1 1 lim = 1 lim = ¡1 + x x!0 x!0¡ x 4 Only for math 140, sections 1 & 020, fall 15 Example 1 - Solution We can see from this graph that the closer x gets to zero from the right the larger (in the positive sense) the function gets, while the closer x gets to zero from the left the larger (in the negative sense) the function gets. The normal limit does not exist since the two one-sided limits have different values. Therefore 1 lim =1 + x x!0 1 lim x!0 x 5 Only for math 140, sections 1 & 020, fall 15 1 lim = ¡1 ¡ x x!0 DNE Definition of Infinite Limits Let f (x) be a function de¯ned on both sides of a, except possibly at a itself. Then lim f (x) = 1 x!a means that the values of f (x) can be made arbitrarily large by taking x su±ciently close to a, but not equal to a. And lim f (x) = ¡1 x!a means that the values of f (x) can be made arbitrarily large negative by taking x su±ciently close to a, but not equal to a. Note: These de¯nitions can be appropriately modi¯ed for the one-sided limits as well. 6 Only for math 140, sections 1 & 020, fall 15 Something important 1. The symbol ∞ is not a number. 2. limx!a f(x) = 1 or ¡1 does not mean that the limit exist. It simply expresses the particular way in which the limit does not exist. So if limx!a f(x) = 1 or ¡1, we can also say that limx!a f(x) DNE. 7 Only for math 140, sections 1 & 020, fall 15 Graphs of infinite limits (e) lim f (x) = lim+ f (x) = lim f (x) = 1 ¡ x!a x!a x!a (f) lim f (x) = lim f (x) = lim f (x) = ¡1 + ¡ x!a x!a x!a Note: The line x=a is called a vertical asymptote of the curve in all these cases. 8 Only for math 140, sections 1 & 020, fall 15 Definition of Vertical Asymptotes The line x = a is called a vertical asymptote of the curve y = f(x) if f (x) has an in¯nite limit (possibly one-sided) at a. In other words, the line x = a is a vertical asymptote if at least one of the following statements is true: lim f (x) = 1 x!a lim f(x) = ¡1 x!a 9 lim f (x) = 1 ¡ x!a lim f(x) = ¡1 ¡ x!a lim+ f (x) = 1 x!a lim f(x) = ¡1 + x!a Only for math 140, sections 1 & 020, fall 15 Example 2 6 Let f(x) = x2 , evaluate limx!0 f(x), and ¯nd the vertical asymptote(s) of f(x). Solution: As x becomes close to 0, x2 also becomes close to 0, and 6/x2 becomes very large. In fact, the values of the function 6/x2 can be made arbitrarily large by taking x close enough to 0. Therefore 6 lim 2 = 1 x!0 x and x=0 is a vertical asymptote of f(x). 10 Only for math 140, sections 1 & 020, fall 15 Example 2 - Solution Note: Here is a quick sketch of the graph to verify the limits 6 lim 2 = 1 x!0+ x 6 lim 2 = 1 x!0¡ x 6 lim =1 x!0 x2 11 Only for math 140, sections 1 & 020, fall 15 Example 3 Let f(x) = ¡4 . x+2 (a) Evaluate each of the following limits lim + f(x) x!¡2 lim ¡ f (x) x!¡2 lim f(x) x!¡2 (b) Find the vertical asymptote(s) of f (x). 12 Only for math 140, sections 1 & 020, fall 15 Example 3 - Solution Solution: (a) If x is close to ¡2 but larger than ¡2, then the denominator x + 2 ¡4 is a small positive number . So the quotient x+2 is a large negative number. Thus ¡4 lim = ¡1 x!¡2+ x + 2 Likewise, if x is close to ¡2 but smaller than ¡2, then x + 2 is ¡4 a small negative number . So the quotient x+2 is a large positive number . Thus ¡4 lim ¡ =1 x!¡2 x + 2 Since limx!¡2+ ¡4 x+2 6= limx!¡2¡ ¡4 , x+2 ¡4 lim DNE x!¡2 x + 2 13 Only for math 140, sections 1 & 020, fall 15 Example 3 - Solution ¡4 lim + = ¡1 x!¡2 x + 2 ¡4 lim ¡ =1 x!¡2 x + 2 ¡4 lim DNE x!¡2 x + 2 (b) The vertical line x = -2 is a vertical asymptote of f (x). 14 Only for math 140, sections 1 & 020, fall 15 Example 4 Let f(x) = 3x¡1 . (x¡5)3 (a) Evaluate each of the following limits lim f(x) + x!5 lim f (x) ¡ x!5 lim f(x) x!5 (b) Find the vertical asymptote(s) of f (x). 15 Only for math 140, sections 1 & 020, fall 15 Example 4 - Solution Solution: (a) Since (x ¡ 5)3 ! 0+ and (3x ¡ 1) ! 14 as x ! 5+ , we have lim + x!5 3x ¡ 1 =1 3 (x ¡ 5) Likewise, Since (x ¡ 5)3 ! 0¡ and (3x + 1) ! 14 as x ! 5¡ , we have 3x ¡ 1 lim = ¡1 ¡ (x ¡ 5)3 x!5 Since limx!5+ 3x¡1 (x¡5)3 6= limx!5¡ 3x¡1 , (x¡5)3 3x ¡ 1 DNE x!5 (x ¡ 5)3 lim 16 Only for math 140, sections 1 & 020, fall 15 Example 4 - Solution Tips: To determine if (x ¡ 5)3 ! 0+ or 0¡ as x ! 5+ , we can choose a test number which is close to 5 but greater than 5 for x and see if (x ¡ 5)3 is positive or negative. For example, let x = 5:1 and we have (x ¡ 5)3 = (5:1 ¡ 5)3 = (0:1)3 > 0: So (x ¡ 5)3 ! 0+ as x ! 5+ . Note: Only when the denominator of the function approaches 0 as x approaches the given point we can use this method to determine the sign of the denominator. 17 Only for math 140, sections 1 & 020, fall 15 Example 4 - Solution Here is a quick sketch of the graph to verify the limits (b) x=5 is a vertical asymptote of the function f (x). 18 Only for math 140, sections 1 & 020, fall 15 How to evaluate an infinite limit Tips g(x) , if h(x) ! 0 and g(x) approach a h(x) nonnegative number as x ! a, then limx!a f(x) has an in¯nite limit at x = a. Given a function f(x) = Case 1: If h(x) ! 0+ and g(x) ! +c ( or ¡c ), where c > 0, as x ! a, then limx!a f(x) = 1 ( or ¡ 1). Case 2: If h(x) ! 0¡ and g(x) ! ¡c ( or +c ), where c > 0, as x ! a, then limx!a f(x) = ¡1 ( or 1). 19 Only for math 140, sections 1 & 020, fall 15 Example 5 Find the vertical asymptotes of f (x) = tan x. Solution: sin x Because tan x = cos x there are potential vertical asymptotes where cos x = 0. We notice that cos ( /2)=0. In fact, since cos x 0+ as x ( /2)- and cos x 0- as x ( /2)+, whereas sin x is positive when x is near /2, we have lim ¡ tan x = 1 and lim + tan x = ¡1 ¼ ¼ x!( 2 ) x!( 2 ) This shows that the line x = /2 is a vertical asymptote. 20 Only for math 140, sections 1 & 020, fall 15 Example 5 - Solution Similar reasoning shows that the lines x = (2n + 1) /2, where n is an integer, are all vertical asymptotes of f(x) = tan x. The graph confirms this. 21 Only for math 140, sections 1 & 020, fall 15 How to find vertical asymptotes Tips g(x) The line x = a is a vertical asymptote of the function f (x) = h(x) if h(x) ! 0 and g(x) remains nonegative as x ! 0. g(x) In order to ¯nd the potential vertical asymptotes of , h(x) just ¯nd the values x where h(x) = 0 and check if g(x) 6= 0. 22 Only for math 140, sections 1 & 020, fall 15 Example 6 Find the vertical asymptotes of the function 2x4 + 5 f(x) = 9 ¡ x2 23 Only for math 140, sections 1 & 020, fall 15 Example 6 - Solution Solution: There are potential vertical asymptotes where the denominator (9 ¡ x2 ) is equal to 0. Since (9 ¡ x2 ) = (3 + x)(3 ¡ x) = 0 when x = ¡3 or 3 (and the numerator (2x4 + 5) is not 0), x = ¡3 and x = 3 are vertical asymptotes of the function. 24 Only for math 140, sections 1 & 020, fall 15 ...
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