S(x) Section 1.5 infinite Limits. lim f(x)=-m x --," -3 + Jkx) - x ~
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1 8 Chapter Limits and Their Properties Section.5 infinite Limits -. f(x)- (x-) As x approaches from the left, x - is a small negative number. So, lim f(x) : -m As x approaches from the right, x - is a small positive number. So, lim f(x] = oo x---~ + - ~tx) - x - As x approaches from the left, x - is a small negative number. So, lim f(x) = m. As x approaches from the right, x - is a small positive number. So, lim [(x) = x--> + 3. f(x)- (x - ) As x approaches from the left or right, (x - ) is a small positive number. So, lim f(x) = lim f(x) = ~. x--~ + x-~- Jkx) - x ~ - 9 As x approaches from the left or right, (x - ) is a small positive number. So, lim f(x) = lim f(x) = -~. x--> - x lim ~ = m x-~- + --~x - lim ^x = oo x-,--_.~ax - 6. lim - oo x-~- + x + lim ~ x--,,--x + ~X 7. lim tanm = _m x~- + ~X lim tan- = m ~X 8. lim sec-- = m ;rt x lim sec-- =-~ x~-- x :(x) lim f(x) = oo lim f(x)=-m x --," -3 + x 0. f(x) - x= _ 9 x S(x) lim f(x)= lim f(x) = m x --~ Brooks/Cole, Cengage Learning
2 Section.5 nfinite Limits 85 X. f(x) - x _ 9 x s(-) lim f(x) : ~ x-# -3- lim f(x)=-~ x-# , X = sec-- 6 x s(-) lim f(x)=-oo lira f(x) = oo lim -=- = oo = lim X x-#o + X z Therefore, x = 0 is a vertical asymptote.. lim-- - oo x-#+(x- )3 5. lim-- oo )3 Therefore, x = is a vertical asymptote. X x lim z - mand lim x~ - m x -#-~- x - x -#-~ + - Therefore, x = - is a vertical asymptote. X x-#- X -- X and lim x-#+ x - - m Therefore, x = is a vertical asymptote. 6. No vertical asymptote because the denominator is never zero. 7, No vertical asymptote because the denominator is never zero. 8. lim h(s} : -oo and lim h(s~ =~o. s-#-5-~ ~ s-#-5 + Therefore, s = -5 is a vertical asymptote. lim h(s} = -oo and lim h(s} s-#5-~ ~ s-#5 + Therefore, s = 5 is a vertical asymptote. x - 9, lim = m x-#~+(x- )(x + ) x = - lim = -m,<-(x- )(x + 0 Therefore, x = is a vertical asymptote. x - lim = ~ x-#-,+(x- )(x + ) X -- lim = -~ x-#--(x- )(x + ) Therefore, x = - is a vertical asymptote. 00 Brooks/Cole, Cengage Learning
3 86 Chapter Limits and Their Properties +x +x 0. lim - lim = ~ x ~-+0-x( - x) x~( x) Therefore, x = 0 is a vertical asymptote. lira +x - m x-+l- X( - X) +x lim x~( x) -- x.+l + Therefore, x = is a vertical asymptote.. lim(-) -m= lim(-7 ) Therefore, t = 0 is a vertical asymptote. (/)x 3 - x - x l X(X - x - 8). g(x) = 3x ~ - 6x - = 6 X -- x - 8 = -x, x, -, 6 No vertical asymptote. The graph has holes at x = - and x = f(x)= x + x- (x + )(x-) Vertical asymptotes at x = - and x =. (x + 3)(x- ) x(x - )(; - 9) - x :/: -3, x(x - 3) Vertical asymptotes at x = 0 and x = 3. The graph has holes at x = -3and x =. - x3+l (x + l)(x - x + l) x+l x+l has no vertical asymptote because lxi_+m_,f(x ) : lim(x - x + ): 3. x-+- \ The graph has a hole at x = -. x - (x+)(x-) 6. h(x)= x +x +x+ = (x+)(x +) has no vertical asymptote because - lim h(x) = lira ~ -. (x- s)(x + 3) i-~x + 7. f(x) = x-~53. (x-5)(x +) + No vertical asymptote. The graph has a hole at x = 5. ~. h(t) = t(t- ) (t- )(t + )(t + ) t = (t + )(fl + ) # Vertical asymptote at t = -. The graph has a hole at t=. sin ~rx 9. f(x) = tan a-x = ~ has vertical asymptotes at COS 7/ X x - n+, n any integer. 30. f(x) = sec a x = ~ has vertical asymptotes at COS 7gX x - n+, n any integer. t 3. s(t) =-- has vertical asymptotes at t = nzc, n a sin t nonzero integer. There is no vertical asymptote at t = 0 since lim t--~-- =.,-+0sin t tan 3. g(o) - - has vertical asymptotes at 0 0 cos 0 0 = (n + )zc - ~r+ nn-, n any interger. There is no vertical asymptote at 0 = 0 because limtan0 =. o-+0 0 x lira - lim(x-) =- -5 Removable discontinuity at x = - The graph has a hole at x = Brooks/Cole, Cengage Learning
4 Section.5 nfinite Limits 87 x - 6x lim = lim(x-7) =-8 x-+-a X + x-+-l" - Removable discontinuity at x = - X lim - oo x-+- + X + X + lim _ x-+-l- x + Vertical asymptote at x = - 8 x+3 x+3 3. lim 6) lim x-+-~-(x + x- x-+-~-(x + 3)(x- ) = lim - x-+-3-x - 5. lim 6x = +x- lim (3x-)(x+l) x-+-(/) + x - x - 3 x-+-o/)+(x - 3)(x + ) 3x - 5 lim - x-+-(/) + x lim., x- = lim-~ _ x-+l(x + )(x- ) x-+ix + 6. limx- _ x-+3 X lim sin(x + ) _ x-+-i x + Removable discontinuity at X = lim-- = oo x-+o + sin x lim - oo x-+(~/) + cos x 5. lim l~x = lim(~x sin x)=0 x-+or CSC X ~-+~r 5. lim(x x-+o cot + x ) : x-+o~lim~(x + )tan x] = lim - m x-+-i + x lim--- ~o x~l-(x -- ) X 39. lim - oo x-+ + x - +x 0. lira - oo x-+l + -- X x. lim-- _ oo x-++(x -- ) X. lim - x-+-x lim x sec(rex) : ~ and lim x sec(rex) x-+(l/)- x-+(/) + Therefore, lim x see(rex) does not exist. x-+(/) 5. lim x tan rex = m and x-+(/)- lim x tan rex = -oo. Therefore, X-+(/) + lira x tan rex does not exist. x x 55. f(x) - + x + l _ + x + l x 3 - (x-)(x +x+i) lim f(x) lim x-+l + x-+l + x Brooks/Cole, Cengage Learning
5 Chapter Limits and Their Properties 56. f(x) - x + x q- _ - + x x + x + limf(x) = lim(x-) = 0 x-~-l- 6. No, it is not true. Consider p(x) = x -. The function x - p(x) f(x) - x- x- has a hole at (, ), not a vertical asymptote. lira f(x) = -~ mo - (vv lim m = lim k 66. P=-- V m0 0,3 lim--k = k(~) = ~ v-+o + V (n this case you know that k > 0. ) 67. (a) r = 50re sec ~ = 00 ~ if/see 6 3 ~X 8 lim f(x) = x-~ + (b) r = 50~ see 7~ = 00n" if/see 3 (c) lim E50n-sec 0~ = ~ o~(x/)- 68. (a) r = -,/65-9 (7) = 7-~ ff/sec (5) 3 ft/sec (b) r = x/65-5 = ~ A limit in which f(x) increases or decreases without bound as x approaches c is called an infinite limit, m is not a number. Rather, the symbol lim/(x) : oo X ~C says how the limit fails to exist. The line x = c is a vertical asymptote if the graph off approaches + m as x approaches c. 6. One answer is x -3 = x-3 f(x) : (x- 6)(x + ) x - x- " 6. No. For example, f(x) = ~ has no vertical 63. asymptote ! [ ~ -x x (c) lira - ~ x-~5- ~/65 - x Total distance 69, (a) Average speed = Total time d 50= (d/x) + (d/y) (b) 50 = xy y+x 50y + 50x = xy 50x = xy - 50y 5x =y x - 5 Domain: x > 5 50x : y(x- 5 / x 3O O 50 6O Y O.857 5x (c) lim x-~5 + ~ - 5 oo As x gets close to 5 mi/h, y becomes larger and larger Brooks/Cole, Cengage Learning
6 Section.5 nfinite Limits (a) x f(x) ~0 ~0 ~ "~" ~.5 (b) -0,5 x - sin x lim- - 0 x->o + X x f (x) ~0 ~ (c) -0.5 x - sin x lim ~ x ~O + X =0 x f(x) O (d) x - sin x lira x (/6) x-->o + x f(x) x - sin x lim- = ~ x_>o + X x - sin x For n > 3, lim-- X n 00 Brooks/Cole, Cengage Learning
7 90 Chapter Limits and Their Properties 7. (a) A = -bh ---rz0 = ~(0)(0tan O)- ~(0)0 = 50tanO (b) Domain: (0, ~/ /(o) loo 0.5 o (c) lira A = oo 0-~r/- 7. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 700/ = 850 revolutions per minute. (b) The direction of rotation is reversed. (c) (0cot0)+ (0 cot 0) : straight sections. The angle subtended in each circle is ~"-((@- (d) So, the length ofthe belt around the pulleys is 0(~c + 0) + 0(~- + 0) = 30(~ + 0). Total length = 60 cot (rc + 0) Domain: (0, ~) L (e) 50 o (f) lim L = 60~- ~ (All the belts are around pulleys.) (g) liml = oo -~ False. For instance, let x - f(x)- x- or X g(x) = + 7. True 76. False. Let f(x) ~ x ~ 0 [3, x=0. The graph off has a vertical asymptote at x = 0, but f(o) = False. The graphs of y = tanx, y = cotx, y = secxand y = cscxhave vertical asymptotes. 00 Brooks/Cole, Cengage Learning
8 Section.5 nfinite Limits and c = 0. Let f(x) : ~ and g(x) : x--~, lim--~ -_ = ~ and lim. = ~, but x.-~ O x x.-,,. O x lim( ) = lim(x -) x-" OC x ---- ~) x--~ot X ) = Tt: O. Given lira ()-- f x (-- and limg x L: () Difference: Let h(x) = -g(x). Then lim h(x) = -L, and X--~C ~i~m~{f(x) g(x)] = x-~c - lira E ( )+ f x h(x)] () Product: = ~, by the Sum Property. f L > 0, then for ~ = L/ > 0 there exists fil > 0 such that [g(x) - L < L/ whenever 0 < x- c[ <.So, L/ < g(x) < 3L/. Because limf x = then for M > 0, there exists 8 > 0suchthat f(x) > M(/L)whenever x - c < 8. Let 8 be the smaller of 8 and 8. Then for 0 < x - c < 8, you have f(x)g(x) > M(/L)(L/)= M. Therefore limf x~c ( ) x ( g ) = x m The proof is similar for L < 0. (3) Quotient: Let ~ > 0 be given. There exists 8 > 0 such that f(x) > 3L/Z~whenever 0 < x-c < 8 and there exists 8 > 0 such that g(x) - L[ < L/ whenever 0 < x - c[ < 8. This inequality gives us L/ < g(x) < 3L/. Let 8be the smaller of 8 and 8. Then for 0 <[x-c < 8, youhave g(x) 3L/ f(x) < 3L/~ - ~" 79. Given limf x ~, let x~c ( )-- g(x) =. Then j g(x) f(x) = 0 by Theorem Givenx-~f(x)lim-- = 0. Suppose.~limf()x exists and equals L. lira Then, lim _ x~c =-- = 0. :~-~ f(x) lira f(x) L X- ~,C This is not possible. So,x~c ( )limf x does not exist. f( ) -- is defined for all x > 3. Let M > 0 be given. You need 8 > 0 such that f( ) =-- > Mwhenever 3 < x < 3+8. Equivalently, x -3 < --whenever M ]x-3 <8, x>3. So take 8 = --:---. Then for x > 3 and M x 3<8, -- >-= Mandso f(x] > M. x-3 8 Therefore, ~i~m~ g(x) = O f(x) : ~ is defined for all x < 5. Let N < 0 be given. You need 8 > 0 such that f(x) : ~ x-5 < N whenever x < x < 5. Equivalently, x - 5 > --whenever x - 5 < 8, x < 5. Equivalently, < --- N whenever N x-5 < 8, x < 5. So take 8 = _l. Note that 8 > 0becauseN < 0.For x-5 < 8and N x < 5,,~_, > -- = -N, and - (5 x-5 x-5[ <N ix-~l 00 Brooks/Cole, Cengage Learning
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