solution(common) - 1. [30 pt] Let f(m) ={ Edwin/m), :3: (3)...

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Unformatted text preview: 1. [30 pt] Let f(m) ={ Edwin/m), :3: (3) Using an e — 6 argument, show f is not continuous at .1: = 0 when a = 0 . sol.) I A when a = 0 , ‘ __ sin(1/:1:) ifa:>0, f(m)‘{0 ifng. we’ll show that discontinuity of f at 0, that is, 350 = %,V6 > O, 3.7:; 3.1;. [$5 —— 0| < 6 and |sin(1/m5) —- 0| 2 60 = (1) In particular, V6, 3n 6 N s.t. < 6.(by Archemedes’ thm. ) %(2n — 1) we can take 1:5 = m, which is depending on any small 6 , and we also 2 get |sin(1/x5) — 0| = |sin(§(2n —— 1)) — 0| = 1 Z :7 = so . Thus (1) holds. So, f is not continuous at a: = O. . wegawwmneeeaHessea2m%wwrsssaa «m see was we. JHWHM4QNfifi%%ififl%fifidEe—6%¢Wfififlfl% we see-:— Haiti‘s, 9mg see-es we. 5‘3,sin(1/w) 7} 73% a7] Ira-eon see #2121 %§}D}% we :4 exam %°l°fl we a ~%%HNEHH%flflflfl%flfiflflfifa£izfifireflfifli Ix E 94:32} asaTs —’r‘— g. (b) Determine all a that make f continuous at a: = 0 . _ m“sin(1/m), an > 0, sol.) Suppose f(ac) — { O, x S 0 At first, obviously limznO— f = 0 and f (0) = 0 . Since f is continuous at 0, limz_.0+ x“ sin(1/av) must be zero. To satisfy that , we’ll show that the unknown value a must be lager than zero. Otherwise, the limit limm_,0+ x“ sin(1/x) diverges. We consider that a is divided into three cases. is continuous. Case1:a > 0 ) We have |sin(:-:-)| < 1 for all :1: 7E 0. Since 1 —w“ 3 9:“ sin(;) 5 1:“ for all a: > 0, By the sandwich thm. ,limz_,0+ x“ sin(1 = 0. L Case2 : a = 0 ) Through the previous problem (a), we have already considered. We have 1' a ' 1 = 1' ' 1 '11 t . Elgar s1n( /m) mirng s1n( /:1:)osm a es Case3: a < 0 ) We have Since |sin(%)| < 1 for all a: 0,‘ limzq0+ as“ sin(1/a:) = 00. Thus, f is continuous at 0 only in Casel 2 a > 0 . (EH3 773) 78%" a > 0 E’l-é— $513 57%. %°lll-7§7lel ERR“; 7E. (c) Determine all a that make f differentiable at as =0. sol.) Suppose that f is differentiable at 0. the“ f( ) 1(0) I __ . a: —‘ . f (0) — along) x _ 0 ex1st . Obviously, m—vO“ (II—0 3-90“ (II-0 . By the way, —— a . —- lim M = lim M9 = lim xa‘l sin(l) must be zero , :c—+0+ ‘ (B — 0 mfi0+ (I! — 0 z—>0+ (1: since f’ (0) exist . To satisfy that , we’ll show that the unknown value a > 1. Otherwise, the limit limzq0+ ma‘l sin(1/:r) diverges. We consider that a is divided into three cases. Case1:a >1 ) We have lim :1: =0 |sin(%)l < 1 for all a: 75 0. Since —a:"’1 S 3‘14 Sing) 5 (pa—1 for all a: > 0, limm_,o+ era—1 sin(1/:c) = O by the sandwich thm. Case2: a = 1 ) We have lim 27“"1 sin(1/a:) = lim sin(1/:c) oscillates. x—90+ x—+0+ Case3: a < 1 ) We have . _ . 1 11m x“ 1 = hm 1 -oo. z—>0+ :1:—->0+ (12”a+ Since Isin(%)[ < 1 for all a: aé 0, lim$_,o+ 11:“‘1 sin(1/:z:) 7.: 00. Thus, f is differentiable at 0 only in Casel : a > 1 . ($1431 7%) 7d? a > 1 E1173: “£319; 575}. %°l-T'Jr7§7/lxl Elil‘i'l; 77g- ((1) Determine all a that make f’ (as) continuous at :1: = 0 . sol.) Suppose that ama'l sin(1/m) + a3“ cos(1/m)(—1/$2), f'a) ={ ={ am“*‘sin<1/x>— wa-Zcosa/z); x > o, 0, 0, is continuous at an = 0. At first, obviously limmfio— f’ = O and f’ (0) = 0 . Since f ’ is continuous at 0, limx_.0+ ama‘l sin(1/x) — #4 cos(1 /x) must be zero. To satisfy that , we’ll show that the unknown value a‘ must be lager than two.(i.e. a > 2 ) Otherwise, the limit limx_,0+ {acca'1 sin(1/av)—:1:“‘2 cos(1/m)} diverges. We consider that a is divided into cases. Case: a > 2 ) We have 11m x“_1 = O art—90+ lim (ca—2 = 0 m-»0+ _ 1 |sm(;)| < 1 for all :1: 74 0. 1 lcos(;)| < 1 for all m 95 0. By the sandwich thm. , limw_,0+ ama'l sin(1/:1:) - wa'2 cos(1/:1:) = 0 The other cases are similar with case2, 3 of the previous problems (b), (c), so that lim$_,0+ area-1 sin(1/:1:) — #4 cos(1/:c) diverges in that cases. Thus, f’ is continuous at 0 only for a > 2 . p (as 7%?!) 721%}; > 2 "El-3- B’lfiltfil 5s. %°l3l'7§7l7‘l ERIE 7%. $50 2. [20 Point] (a) [10 Point] -solution 1- 1 + tan29 = sec29 => 1 + (t2)2 = sec20 0 (1‘0) ‘x =9 1+t4=sec20 HE 320“, $3 @QBI klfiam 9% to" 3% EDI 2M1 EIII 91-:- -solution 2- tan(7r-9) = ~1— 2 -ta.n0=tT sec20 d0 =— 2tdt 0 (1,0) x —2t _ ~2t $6030 — 1 + t4 ==> 1+ (— t2)2 = sec20 1 + tan29 = sec20 => 1 +t‘1 = sec20 0 0g 3E0", 5?: E'QBI MESH! 93% I toil 33E l.‘g‘OI gAI EIII Eur-3- (b) [10 Point] -solution 1- 2t _ 2(1-3t‘) ‘ (1+t4)2 position (0,?) ? (0, 7/1?) maximum speed 2! 0|»?— 5183’ -solution 2- solution 101W L?’—§Oi| fi—‘F—fiiOi EDH. 51-i- §¢E solution 1 Bi DéOI 73%. I critical point i fiiLiEi i%-‘u’- — '3 ——AJ£ '3 tt((°9+. -SIH£)4 I 511%“: et(mg_£ +S7Ht) t J31 ; d‘fld’a : guest—mm) __ (0ch Hm A? C STL +§3K—- o+‘ :— a '5: J. c. z _- M A% £‘E1: (03% _$.(n:r;_: o _\ ‘ p ) _ A» _ l W“) “M11” ~ Whit) (10—) 1(1)) ‘&(£) -' 93"”; e')(.,( -0) : PDQ + 93"” (§ ‘23 ) &(Jb A1? 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[5pt](b) Find the maximum of f (x) . [Solution] 1 i1“ 1 (a) f(x)=x7:=efi "” 1 ' .... .. x. f'(x)=e$ (—lnx) = * 573, “15' §¥3F5 27% E5 373(ul~‘€:% (b)°lV‘i @1015 %%%i%) To find extreme values, let f '(x)=0 when x > 0. 1 Since x$¢0 if x>0, 2-lnx=0. .'.x=e In addition, 1 l limx"; =1imeJ? and lim—=-oo. x90 x—)0 x—)0 x 1lnx slimeW =0 ---- --(2) x—)0 * 4x3, 7411:} we] gamut m eagle 27s. 1 1 lnx 11m x5 = lim e7: x—ND x->oo and firm—13 =lim =lim L0 x—WDJ; x—No 1 ix-)=n,\/;— L. 2J§ l x —l—lnx 1ime~5 =e° =1 ---- ~(3) X —}ao * 4%], 7%} 3l7§°l Ealflbl 3l7§-% ‘3}101‘3 2’3. Hence, we can see that y =1 is the only asymptote from the table ---- * 2’9, €347} 5391011. x=0. y=1) 07E}. 1 2 .1. (b) f(ez) =(e2)‘le—2 = (62 )2 = e2 is the global maximum value of f (x) ---- * 573, 73347} %fl% 73-?- ”eljéql “til 07%;] EE 2%] $1;— 3421. © TN ATstW "gram HM KN Ufa-fl +0 the, cmté, A: \0—~U- 4’ —\\ 3% \I l’wkfi“ 77:9 Volume :59 TN. 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solution(common) - 1. [30 pt] Let f(m) ={ Edwin/m), :3: (3)...

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