. (h - 1) (h - 1) [(h - 1) + 1] [(h - 1)2 - l(h - 1) J 1 O h = lim = lim. f is continuous from the right at 3

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1 144 D CHAPTER LMTS AND DERVATVES EXERCSES 1. (a) (i) lim f () = 3 (ii) lim f() = (iii) lim f() does not eist since the left and right limits are not equal. (The left limit is -.) (iv) lim f() = (v) lim f() = 00 (vi) lim f() = O (vii) lim f() = 4 (viii) lim f() = oo oo (b) The equations ofthe horizontal asmptotes are = -1 and = 4. (c) The equations of the vertical asmptotes are = 0 and =. (d) f is discontinuous at = -3, 0,, and 4. The discontinuities are jump, infinite, infinite, and removable, respectivel.. lim f() = -, lim f() = 0, lim f() = 00, oo oo lim f() = -00, lim f() =, f is continuous from the right at Since the eponential function is continuous, lim ex - = e 1-1 = eo = l o , T =- : : =-3: S..c,' mce ranona functions are contmuous, nn X (3) - 3 = 1 = o.. (+3)(-3) lim = lim -- = --- = - = ( + 3)( - 1) lim - = - 00 since as and 3 < 0 for 1 < < l+x lim (h - 1) + 1 = lim (h - 3h + 3h - 1) + 1 = lim h - 3h + 3h = lim (h _ 3h + 3) = 3 h-----+o h h-----+o h h-----+o h h-----+o " Another solution: Factor the numerator as a sum of two cubes and then simplif.. (h - 1) (h - 1) [(h - 1) + 1] [(h - 1) - l(h - 1) + 1 J 1 O h = lim = lim h ~ ~ h-----+o h h-----+o h = lim [(h - 1) - h + J = = 3 h-----+o 8. lim t - 4 = lim (t + )(t - ) = lim t t-----+t 3-8 t-----+(t-)(t +t+4) t-----+t +t := lim ( vir)4 = CXJ since (r - 9) ----> 0 as r ----> 9 and -( vr) 4 > 0 for r i- 9. r r - 9 r - 9

2 CHAPTER REVEW D m, v == l' 1m 4 - v == li 1m - 1 == 1 v vl v (4-v) v lim u -1 == lim (u + 1)(u -1) == lim (u + 1)(u+ 1)(u -1) == lim (u + 1)(u+ 1) == () == ~ u--+l u 3 + 5u - 6u u--+l u( u + 5u - 6) u--+l u( U + 6)(u - 1) u--+l u( U + 6) 1(7) 7 1. lim JX+"6- == lim [JX+"6-, JX+"6+] == lim (JX+"6) _ ( - 3) VX X --+3 ( - 3)( VX ) ( --6). -(-3)(+) == lim == lim == lim X (X-3) (+6+) --+3X (X-3) (+6+) X--+3X (X-3)(+6+) == lim - ( + ) 5 _ ~ X--+3X (v+6+) 9(3+3) Since is positive, # == l ==. Thus, 14. Since is negative, # == l == -. Thus, lim ~ == lim ~/# == lim Vl- 9/ == l- 0 ==! X--+CX) -6 X--+CX) (-6)/ X--+CX) -6/ -0 lim ~ == lim ~/# == lim Vl- 9/ == Vl- 0 ==_! X--+-CX) -6 X--+-CX) (-6)/(-) X--+-CX) -+6/ Let t == sin. Then as '(-, sin , so t Thus, lim n(sin ) == lim n t == -00. X--+K- t--+o (1- -4)/4 l/ 4-/ lim == lim == lim == == - == X--+-CX) X--+-CX) ( ) / 4 X--+-CX) 5/ 4 + 1/ X+-X. +4X+l+X] +4+l)- 17. lim (v+4+l-)== lim [VX VX == lim ( X--+CX) X--+CX) 1 Y X X--+CX) V == lim (4 + 1)/ X X--+CX) ( )/ [divide b = # for > 0] == lim 4 + l/ 4+0 ==~== X--+CX) V + 4/ + l/ + 1 V Let t == - == (1 - ). Then as X- t , t , and lim e == lim e == O. --+CX) t --+- CX) 19. Let t == l/. Then as , t , and lim tan- 1 (l/) == lim tan- 1 t == ~. --+o+ t--+cx) 0. lim (_1_ + 1 ) == lim [_1_ + 1 ] == lim [ ] --+l l - ( - l)( - ) --+l ( - 1)( - ) ( - 1)( - ) u [ - ] i == ~ ( - 1)( - ) == ~ - == 1 - ==

3 146 D CHAPTER LMTS AND DERVATVES 1. From the graph of = (cos )/, it appears that = 0 is the horizontal asmptote and = 0 is the vertical asmptote. Now 0 :::; (cos ) :::; 1 =} ~ < cos X < ~ 0 < cos ' X < 1 B l' 0 0 d - - =} ' ut fl = an X X-+±()() 1. cos" lim - = 0, so b the Squeeze Theorem, lim --- = O. X-+±()() X-+±()() ) \ 6 Thus, = 0 is the horizontal asmptote. lim cos = 00 because cos'' -7 1 and -7 0 as -7 0, so X = 0 is the -+O vertical asmptote.. From the graph of = f () = vi + X V -, it appears that there are horizontal asmptotes and possibl vertical asmptotes. To obtain a different form for f, let's multipl and divide it b its conjugate. Now it () = (vi + X + _ vi _ ) vi + X + + vi - X =...;...(X-;:=:::;::+===X==+==1::::::-) (~X:::::::; ;:::::::-==:::::::) V + X + 1+ V - X vi + X + 1+ vi X vi + 1+ vi - X f () li + 1 1trn 1 = im X-+()() X-+()() v + X + 1+ vl - X = lim + (/) X-+()() V + (/) + (/ ) + V - (/) [since R = for > 0] , so = 1 is a horizontal asmptote. For < 0, we have R = l = -, so when we divide the denominator b, with < 0, we get Therefore, 1 rm f() 1 ~-oo u + ~-oo vl + X + + vl - = un =-1 -(1 + 1), X u = im ~oo - +(/). [V + (/) + (l/ ) + V (/) ] so = -1 is a horizontal asmptote. The domain of f is (-00,0] U [1, (0). As -7 0-, f () -7 1, so = 0 is not a vertical asmptote. As -7 1+, f () -7!3, so X = 1 is not a vertical asmptote and hence there are no vertical asmptotes..:» ' -4

4 CHAPTER REVEW D Since - 1 :::; f() :::; for 0 < < 3 and lim ( - 1) == 1 == lim, we have lim f() == 1 b the Squeeze Theorem. -+l -+l -+l 4. Let f() == -, g() == cos(1/) and h() ==. Then since cos(1/) :::; 1 for i- 0, we have f () :::; g() :::; h() for i- 0, and so lim f() == lim h() == 0 =? lim g() == 0 b the Squeeze Theorem. -+O -+O -+O 5. Given E > 0, we need 6 > 0 such that if 0 < - 1 < 6, then 1(14-5) - 41 < E. But 1(14-5) - 41 < E q < E q < E q - 1 < E15. SOifwe choose 6 == E15, then 0 < - 1 < 6 =? '(14-5) - 41 < E. Thus, lim (14-5) == 4 b the definition of a limit Given E > 0 we must find 6 > 0 so that if 0 < - 01 < 6, then {X - 01 < E. Now 1{X - 01 == {X < E =? 3 < E 3. l == 1{X SOtake 6 == E 3. Then 0 < - 01 == l < E 3 =? {X - 01 == {X == {M < ~ == E. Therefore, b the definition of a limit, lim {X == O. -+O 7. Given E > 0, we need 6 > 0 so that if0 < - 1 < 6, then (-) < E. First, note that if - 1 < 1, then -1 < - < 1, so 0 < - < =? -11 <. Now let 6 == min {E/, }. Then 0 < - 1 < 6 =? (-) == ( - ) ( - 1) == < (E ) () == E. Thus, lim ( - 3) == - b the definition of a limit Given M > 0, we need 6 > 0 such that if 0 < - 4 < 6, then /v' - 4 > M. This is true q v' - 4 < 1M q - 4 < 41M. So if we choose 6 == 41M, then 0 < - 4 < 6 =? /' - 4 > M. So b the definition of a limit, lim (/' - 4) == (a) f() == ve if < 0, f() == 3 - if 0 :::; < 3, f() == ( - 3) if > 3. (i) lim f() == lim (3 - ) == 3 (ii) lim f() == lim ve == 0 -+O+ -+O+ -+O- -+O (iii) Because of (i) and (ii), lim f () does not eist. (iv) lim f() == lim (3 - ) == 0 -+O (v) lim f () == lim ( - 3) == 0 (vi) Because of (iv) and (v), lim f() == O (b) f is discontinuous at 0 since lim f () does not eist. (c) -+O 3 f is discontinuous at 3 since f (3) does not eist. 30. (a) g() == - if 0 :::; :::;, g() == - if < :::; 3, g() == - 4 if3 < < 4, g() == tt if ~ 4. ) Therefore, lim g() == lim ( - == 0 and lim g() == lim ( - ) == O. Thus, lim g() == 0 == 9 (), so 9 is continuous at. lim g() == lim ( - ) == -1 and lim g() == lim ( - 4) == -1. Thus, lim g() == -1 == g(3), so 9 is continuous at 3. lim g() == lim ( - 4) == 0 and lim g() == lim Jr == Jr. ::e Thus, lim g() does not eist, so 9 is discontinuous at 4. But lim g() == tt == g(4), so 9 is continuous from the right at 4.

5 148 D CHAPTER LMTS AND DERVATVES (b) o sin is continuous on JR. b Theorem 7 in Section.5. Since e" is continuous on JR., e s in is continuous on JR. b Theorem 9 in Section.5. Lastl, is continuous on JR. since it's a polnomial and the product e s in is continuous on its domain JR. b Theorem 4 in Section is continuous on JR. since it is a polnomial and JXis continuous on [0,00), so the composition ' - 9 is continuous on { - 9 : O} = (-00, -3] U [3, 00). Note that - -loon this set and so the quotient function g() = ~ - is continuous on its domain, (-00, -3] U [3,00). 33. f() == is a polnomial, so it is continuous on [-, -1] and f( -) == -10 < 0 < 1 == f( -1). So b the ntermediate Value Theorem there is a number c in (-, -1) such that f (c) == 0, that is, the equation == 0 has a root in (-, -1). 34. f() == e- - is continuous on JR. so it is continuous on [0,1]. f(o) == 1 > 0 > l/e - 1 == f(l). So b the ntermediate Value Theorem, there is a number c in (0,1) such that f(c) == O. Thus, e- - == 0, or e- ==, has a root in (0,1). 35. (a) The slope of the tangent line at (, 1) is lim f() - f() == lim == lim 8 - == lim -( - 4) == lim -( - )( + ) X----7 X - X----7 X - X----7 X - X----7 X - X----7 X - == lim [-( + )] == - 4== -8 X----7 (b) An equation of this tangent line is - 1 == -8( - ) or == For a general point with -coordinate a, we have m == lim /(1-3) - /(1-3a) == lim (1-3a) - (1-3) == lim 6( - a) X----7a X - a X----7a (1-3a)(1-3)( - a) X----7a (1-3a)(1-3)( - a) == lim 6 6 X----7a (1-3a)(1-3) (1-3a) For a == 0, m == 6 and f(o) ==, so an equation of the tangent line is - == 6( - 0) or == 6 +. For a == -1, m == ~ and f( -1) == ~, so an equation of the tangent line is - ~ == ~( + 1) or == ~ + i. 37. (a) 8 == 8(t) == 1 + t + t /4. The average velocit over the time interval [1, 1 + h] is a v e 8(1 + h) - 8(1) 1 + (1 + h) + (1 + h)/4-13/4 10h + h 10+ h V == (1 + h) - 1 == h == 4h 4 So for the following intervals the average velocities are: (i) [1,3]: h ==, Va ve == (10 + )/4 == 3 m/ s (iii) [1,1.5]: h == 0.5, Va v e == ( )/4 ==.65 m/s (ii) [1,]: h == 1, Vave == (10 + 1)/4 ==.75 m/ s (iv) [1,1.1]: h == 0.1, Va v e == ( )/4 ==.55 m/s

6 CHAPTER REVEW l' S (1 + h) - s (1) li 10 + h 10 / (b) When t = 1, t he nstantaneous ve OCt S im h = im-- = - =.5 m s. h~o h~o (a) When V increases from 00 irr' to 50 in", we have ~V = = 50 in", and since P = 800/V, ~P = P(50) - P(00) = = = -0.8 lb/in". So the average rate of change is ~P = -0.8 = lb/in. ~V 50 irr' (b) Since V = 800/P, the instantaneous rate of change of V with respect to P is lim ~V = lim V(P+h)-V(P) = lim 800/(P+h)-800/P = lim 800[P-(P+h)] h~o ~P h~o h h~o h h->o h(p + h)p = lim -800 _ 800 h~o (P + h)p P? which is inversel proportional to the square of P (a) J'() = lim f() - f() = lim (c) 1 ~ X - ~ X -. ( - )( + + ) = lim ,;.----~ ~ X - = lim ( + + ) = 10 ~ (b) - 4 = 10( - ) or = lo = 64, so f() = 6 and a =. 41. (a) ' (r) is the rate at which the total cost changes with respect to the interest rate. ts units are dollars/ (percent per ear). (b) The total cost of paing off the loan is increasing b $100/ (percent per ear) as the interest rate reaches 10%. So if the interest rate goes up from 10% to 11%, the cost goes up approimatel $100. (c) As r increases, C increases. So ff (r) will alwas be positive i o i'

7 150 D CHAPTER LMTS AND DERVATVES 45. (a).f'() = lim f( + h) - f() = lim }3-5( + h) - v:r=s }3-5( + h) + v:r=s h----*o h h----*o h J3="5( + h) + '3-5 =lim [3-5(+h)]-(3-5) =lim -5 = -5 h~o h ( }3-5( + h) +)3-5X) h~o }3-5( + h) +)3-5 )3-5 (b) Domain of f : (the radicand must be nonnegative) ~ 0 ::::} 5 ~ 3 ::::} E (-00, ~ ] Domain of t'. eclude ~ because it makes the denominator zero; E (-00, %) (c) Our answer to part (a) is reasonable because f' () is alwas negative and f L--j if 6 is alwas decreasing (a) As ±oo, f() = (4 - )j(3 + ) , so there is a horizontal asmptote at = -1. As , f() , and as , f() Thus, there is a vertical asmptote at = -3. (b) Note that f is decreasing on (-00, -3) and (-3,00), so f' is negative on those intervals. As ±oo, f' O. As and as , f' =-3 o if 4- (+h) _ 4- (C)f'(X) = lim f(+h)-f() = lim 3+(+h) 3+ = lim (3+)[4-(+h)]-(4-)[3+(+h)] h----*o h h----*o h h----*o h[3+(+h)](3+). ( h h) - ( h h) =hm-' '---' ' h----*o h[3 + ( + h)](3 + ) -7h -7 7 = lim = lim h----*o h[3+ (+h)] (3+) h----*o [3+ (+h)] (3+) (d) The graphing device confirms our graph in part (b). 47. f is not differentiable: at = -4 because f is not continuous, at = -1 because f has a comer, at = because f is not continuous, and at = 5 because f has a vertical tangent. 48. The graph of a has tangent lines with positive slope for < 0 and negative slope for > 0, and the values of c fit this pattern, so c must be the graph of the derivative of the function for a. The graph of c has horizontal tangent lines to the left and right of the -ais and b has zeros at these points. Hence, b is the graph of the derivative of the function for c. Therefore, a is the graph of f, c is the graph of f', and b is the graph of i".

8 CHAPTER REVEW D / (1990) is the rate at which the total value ofus currenc in circulation is changing in billions of dollars per ear. To estimate the value of 0/(1990), we will average the difference quotients obtained using the times t == 1985 and t == L t A == 0(1985) - 0(1990) == == == 16.9 d e 1985 _ an B == 0(1995) - 0(1990) == == == 7 48 Th en 0/(1990) == r O(t) - 0(1990) r--.j A + B == == 44.4 == billi d 11 / t--}ru-90 t _ 1990 r--.j ' on 0 ars ear (a) Drawing slope triangles, we obtain the following estimates: F ' (1950) ~ ~.~ == 0.11, F' (1965) ~ -1 == -0.16, and F' (1987) ~ ~.g == 0.0. (b) The rate of change of the average number of children born to each woman was increasing b 0.11 in 1950, decreasing b 0.16 in 1965, and increasing b 0.0 in (c) There are man possible reasons: n the bab-boom era (post- WW), there was optimism about the econom and famil size was rising. n the bab-bust era, there was less economic optimism, and it was considered less sociall responsible to have a large famil. n the bab-boomlet era, there was increased economic optimism and a return to more conservative attitudes. 51. f() :::; g() ~ -g():::; f() :::; g() and lim g() == 0 == lim -g() a ----+a Thus, b the Squeeze Theorem, lim f () == O a 5. (a) Note that f is an even function since f () == f (- ). Now for an integer n, [n] + [-n] == ti n == 0, and for an real number k which is not an integer, o 3 4 [k] + [-k] == [k] + (- [k] - 1) == -1. So lim f () eists (and is equal to -1) ----+a for all values of a. (b) f is discontinuous at all integers