x 2 k 2 f 99 0 0 because c 99 0 29 a lim x sin x x lim x 1 x 2 3 x 4 5 1 b lim

X 2 k 2 f 99 0 0 because c 99 0 29 a lim x sin x x

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x2k+2;f(99)(0) = 0 becausec99= 0.29. (a)limx0sinxx= limx0(1x2/3! +x4/5!− · · ·)= 1(b)limx0tan1xxx3= limx0(xx3/3 +x5/5x7/7 +· · ·)xx3=1/330. (a)1cosxsinx=1(1x2/2! +x4/4!x6/6! +· · ·)xx3/3! +x5/5!− · · ·=x2/2!x4/4! +x6/6!− · · ·xx3/3! +x5/5!− · · ·=x/2!x3/4! +x5/6!− · · ·1x2/3! +x4/5!− · · ·, xnegationslash= 0;limx01cosxsinx=01= 0(b)limx01xbracketleftbigln1 +xsin 2xbracketrightbig= limx01xbracketleftbigg12ln(1 +x)sin 2xbracketrightbigg= limx01xbracketleftbigg12parenleftbiggx12x2+13x3− · · ·parenrightbiggparenleftbigg2x43x3+415x5− · · ·parenrightbiggbracketrightbigg= limx0parenleftbigg3214x+32x2+· · ·parenrightbigg=3/231.integraldisplay10sin(x2)dx=integraldisplay10parenleftbiggx213!x6+15!x1017!x14+· · ·parenrightbiggdx=13x317·3!x7+111·5!x11115·7!x15+· · ·bracketrightbigg10=1317·3!+111·5!115·7!+· · ·,but115·7!<0.5×103sointegraldisplay10sin(x2)dx1317·3!+111·5!0.310332.integraldisplay1/20tan1(2x2)dx=integraldisplay1/20parenleftbigg2x283x6+325x101287x14+· · ·parenrightbiggdx=23x3821x7+3255x11128105x15+· · ·bracketrightbigg1/20=23123821127+325512111281051215− · · ·,but3255·211<0.5×103sointegraldisplay1/20tan1(2x2)dx23·23821·270.0804
470Chapter 933.integraldisplay0.20(1 +x4)1/3dx=integraldisplay0.20parenleftbigg1 +13x419x8+· · ·parenrightbiggdx=x+115x5181x9+· · ·bracketrightbigg0.20= 0.2 +115(0.2)5181(0.2)9+· · ·,but115(0.2)5<0.5×103sointegraldisplay0.20(1 +x4)1/3dx0.20034.integraldisplay1/20(1 +x2)1/4dx=integraldisplay1/20parenleftbigg114x2+532x415128x6+· · ·parenrightbiggdx=x112x3+132x515896x7+· · ·bracketrightbigg1/20= 1/2112(1/2)3+132(1/2)515896(1/2)7+· · ·,but15896(1/2)7<0.5×103sointegraldisplay1/20(1 +x2)1/4dx1/2112(1/2)3+132(1/2)50.490635. (a)Substitutex4forxin the MacLaurin Series forexto obtain+summationdisplayk=0x4kk!. The radius of conver-gence isR= +.(b)The first method is to multiply the MacLaurin Series forex4byx3:x3ex4=+summationdisplayk=0x4k+3k!. Thesecond method involves differentiation:ddxex4= 4x3ex4, sox3ex4=14ddxex4=14ddx+summationdisplayk=0x4kk!=14+summationdisplayk=04kx4k1k!=+summationdisplayk=0x4k1(k1)!. Use the change of variablej=k1 to show equality of the two series.36. (a)x(1x)2=xddxbracketleftbigg11xbracketrightbigg=xddxbracketleftBiggsummationdisplayk=0xkbracketrightBigg=xbracketleftBiggsummationdisplayk=1kxk1bracketrightBigg=summationdisplayk=1kxk(b)ln(1x) =integraldisplay11xdxC=integraldisplaybracketleftBiggsummationdisplayk=0xkbracketrightBiggdxC=summationdisplayk=0xk+1k+ 1C=summationdisplayk=1xkkC,ln(10) = 0 soC= 0.(c)Replacexwithxin Part (b): ln(1 +x) =+summationdisplayk=1(1)kkxk=+summationdisplayk=1(1)k+1kxk(d)+summationdisplayk=1(1)k+1kconverges by the Alternating Series Test.(e)By Parts (c) and (d) and the remark,+summationdisplayk=1(1)k+1kxkconverges to ln(1 +x) for1< x1.
Exercise Set 9.1047137. (a)In Exercise 36(a), setx=13,S=1/3(11/3)2=34(b)In Part (b) setx= 1/4, S= ln(4/3)38. (a)In Part (c) setx= 1, S= ln 2(b)In Part (b) setx= (e1)/e,S= lne= 1 39. (a) sinh1x=integraldisplay(1 +x2)1/2dxC=integraldisplay parenleftbigg112x2+38x4516x6+· · ·parenrightbiggdxC=parenleftbiggx16x3+340x55112x7+· · ·parenrightbiggC; sinh10 = 0 soC= 0.

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