Infi-1 10

1a

Prove by definition: limx→4x3−3x−x+12x2+x+17=1Proof:Let xn≠4,xn→4Then f(xn)=xn3−3xn−xn+12xn2+xn+17limn→∞2xn2⏟→2⋅42=32+xn⏟→4=2+17=51≠0⟹limn→∞xn3−3xn−xn+12xn2+xn+17=limn→∞xn3−3xn−xn+1limn→∞2xn2+xn+17==151⋅limn→∞xn3⏟→43=64−3xn⏟→3⋅4=12−xn⏟→4=2+1=151⋅51=1⟹limx→4x3−3x−x+12x2+x+17=1

1b

Let f(x)={x2+8x<28x2+2x≥2Prove by definition: ∄limx→2f(x)Proof:Let xn<2,xn→2limn→∞f(xn)=⏟xn<2limn→∞xn2⏟→22=4+8=12Let yn>2,yn→2limn→∞f(yn)=⏟yn>2limn→∞8yn2⏟→8⋅22=32+2=34xn→2⟹f(xn)→12yn→2⟹f(yn)→34⟹By definition: ∄limx→2f(x)

1c

Prove by definition: ∄limx→1sin⁡(1(x−1)2)Proof:Let xn≠1,xn→1xn→1⟹xn−1→0⟹(xn−1)2→0+⟹1(xn−1)2=∞⟹∄limn→∞sin⁡(1(xn−1)2)⟹By definition: ∄limx→1sin⁡(1(x−1)2)

2a

limx→036−x−616xSolution:limx→036−x−616x=limx→0(36−x−6)(36−x+6)16x(36−x+6)==limx→036−x−3616x(36−x+6)=limx→0−116(36−x⏟→36=6+6)=−116⋅12=−1192⟹limx→036−x−616x=−1192

2b

limx→89−x4−x3Solution:limx→84−x3⏟→2=2≠0⟹limx→89−x4−x3=limx→89−xlimx→84−x3=12⋅limx→89−x=12⟹limx→89−x4−x3=12

2c

limx→∞x1/xSolution:limx→∞x1/x=limx→∞eln⁡(x)/xlimx→∞ln⁡(x)x=0⟹limx→∞eln⁡(x)/x=e0=1⟹limx→∞x1/x=1

2d

limx→036+x+9+x+4+xx2+x4Solution:x→0⟹x2+x4>0,x2+x4→0+x→0⟹36+x→36=6x→0⟹9+x→9=3x→0⟹4+x→4=2⟹limx→036+x+9+x+4+x=6+3+2=11⟹limx→036+x+9+x+4+xx2+x4=∞

2e

limx→π2+tan⁡xtan⁡x+2tan⁡xSolution:Let t=−tan⁡xx→π2+⟹t→∞limt→∞−t−t+2−t=limt→∞tt−2−t=limt→∞11−2−ttlimt→∞2−t⏞→0+t⏟→∞=0⟹limt→∞11−2−tt⏟→0=1⟹limx→π2+tan⁡xtan⁡x+2tan⁡x=1

2f

limx→3sin⁡(x−3)sin⁡(9x2−27x)Solution:limx→3sin⁡(x−3)sin⁡(9x2−27x)=limx→3sin⁡(x−3)x−3⏞→1(x−3)(9x2−27x)sin⁡(9x2−27x)9x2−27x⏟→1==limx→3x−39x2−27x=limx→319x=127⟹limx→3sin⁡(x−3)sin⁡(9x2−27x)=127

2g

limx→0x8esin⁡(65x)(1−cos⁡(x2))7sin⁡(x6)Solution:limx→0x8esin⁡(65x)(1−cos⁡(x2))7sin⁡(x6)=limx→0x8esin⁡(65x)(1−cos⁡(x2))7x6sin⁡(x6)x6⏟→1==17limx→0x2esin⁡(65x)(1−cos⁡(x2))limx→0x2⏟→0esin⁡(65x)⏞→0(1−cos⁡(x2)⏟→1)⏟→0=0⋅e0⋅0=0⟹limx→0x8esin⁡(65x)(1−cos⁡(x2))7sin⁡(x6)=17⋅0=0

2h

limx→0(cos⁡x)1/xSolution:limx→0(cos⁡x)1/x=limx→0eln⁡(cos⁡x)/xlimx→0ln⁡(cos⁡x)x=limx→0ln⁡(1+(cos⁡x−1))cos⁡x−1⏟→1⋅cos⁡x−1x=limx→0cos⁡x−1x==−limx→01−cos⁡xx=−0=0⟹limx→0eln⁡(cos⁡x)/x=e0=1⟹limx→0(cos⁡x)1/x=1

2i

limx→∞(1+x2+x)1−x1−xSolution:limx→∞(1+x2+x)1−x1−x=limx→∞eln⁡(1+x2+x)1−x1−xlimx→∞1+x2+x=limx→∞11+11+x⏟→0=1⟹limx→∞ln⁡(1+x2+x)=ln⁡(1)=0limx→∞1−x1−x=limx→∞1−x(1−x)(1+x)=limx→∞11+x=0⟹limx→∞ln⁡(1+x2+x)1−x1−x=0⋅0=0⟹limx→∞eln⁡(1+x2+x)1−x1−x=e0=1⟹limx→∞(1+x2+x)1−x1−x=1

2j

limx→∞(1+x2+x)1−x1−xSolution:limx→∞1+x2+x=limx→∞11+11+x⏟→0=1limx→∞1−x1−x=limx→∞(1−x)(1+x)1−x=limx→∞1+x=∞⟹limx→∞(1+x2+x)1−x1−x=elimx→∞1−x1−x⋅1+x−(2+x)2+xlimx→∞1−x1−x⋅1+x−(2+x)2+x=−limx→∞1+x2+x=−limx→∞1x⏞→0+1x⏞→02x+1⏟→1=−0=0⟹limx→∞(1+x2+x)1−x1−x=e0=1

3a

Does there exist a∈R such thatf(x)={ax=111+e1x−1x≠1is continuous on R?Solution:limx→1f(x)=limx→111+e1x−1Let x→1+⟹x>1⟹x−1→0+⟹1x−1→∞⟹e1x−1→∞⟹1+e1x−1→∞⟹limx→1+11+e1x−1=0Let x→1−x<1⟹x−1→0−⟹1x−1→−∞⟹e1x−1→0⟹limx→1−11+e1x−1=1⟹limx→1−f(x)≠limx→1+f(x)⟹f is not continuous at 1 for any a⟹f is not continuous on R

3b

Find a,b,c∈R such thatf(x)={a+sin⁡(ex)esin⁡xx<0bx=09+x−cxx>0 is continuous on RSolution:esin⁡x≠0⟹a+sin⁡(ex)esin⁡x is continuousx>0⟹9+x−cx is continuous⟹The only problematic point is 0limx→0−f(x)=limx→0−a+sin⁡(ex)esin⁡xx→0−⟹ex→e0=1⟹sin⁡(ex)→sin⁡(1)x→0−⟹sin⁡x→0⟹esin⁡x→e0=1⟹limx→0−a+sin⁡(ex)esin⁡x=a+sin⁡(1)limx→0+f(x)=limx→0+9+x−cx=limx→0+9+x−c2x(9+x+c)Let c≠3⟹9+x−c↛0⟹limx→0+f(x)=∞⟹c=3⟹limx→0+9+x−9x(9+x+3)=limx→0+19+x+3=16f is continuous at 0 if and only if: limx→0−f(x)=limx→0+f(x)=f(0)⟹a+sin⁡(1)=16=b⟹{a=16−sin⁡(1)b=16c=3⟹f is continuous at 0⟹f is continuous on R

4a

Find discontinuities of f(x)=e−1x3Solution:eg(x) is continuous at x0⟺g(x) is continuous at x0⟹Discontinuities of f are discontinuities of −1x3−1x3 is continuous on R∖{0}f(x) is undefined at 0limx→0−f(x)=limx→0−e−1x3⏞→∞=∞⟹0 is an essential (second kind) discontinuity of f

4b

Find discontinuities of f(x)=|x3+x5+x7|x3+x5+x7Solution:x<0⟹x3+x5+x7<0⟹f(x)=−1x>0⟹x3+x5+x7>0⟹f(x)=1f(x) is undefined at 0limx→0+f(x)=limx→0+x3+x5+x7x3+x5+x7=1limx→0−f(x)=limx→0−−x3+x5+x7x3+x5+x7=−1{limx→0+f(x)∈Rlimx→0−f(x)∈Rlimx→0+f(x)≠limx→0−f(x)⟹0 is a jump discontinuity of f

4c

Find discontinuities of f(x)=1+x1+x3Solution:1+x3=(1+x)(1−x+x2)1−x+x2>0f(x) is undefined at −1limx→−11+x1+x3=limx→−111−x+x2=13⟹−1 is a removable discontinuity of f

4d

Find discontinuities of f(x)=sin⁡x|x|Solution:sin⁡x is continuous on R1|x| is continuous on R∖{0}⟹f(x) is continuous on R∖{0}limx→0+f(x)=limx→0+sin⁡(x)x=1limx→0−f(x)=limx→0−sin⁡(x)−x=−1{limx→0+f(x)∈Rlimx→0−f(x)∈Rlimx→0+f(x)≠limx→0−f(x)⟹0 is a jump discontinuity of f

4e

Find discontinuities of f(x)=x2+1−1xSolution:x2+1−1 is continuous on R1x is continuous on R∖{0}⟹f(x) is continuous on R∖{0}f(x) is undefined at 0limx→0f(x)=limx→0x2+1−1x=limx→0x2+1−1x(x2+1+1)==limx→0xx2+1+1=0⟹0 is a removable discontinuity of f

4f

Find discontinuities of f(x)={7+x−3x2−4x≤9x2+81x2−81x>9Solution:Let g(x)=7+x−3x2−4g(x) is undefined on (−∞,−7)⟹g(x) is also undefined for x→−7−⟹Limit of g(x) when x→−7− is also undefined⟹g can be considered continuous at -7, as limx→−7+g(x)=g(−7)=−115∀x≤9:f(x)=g(x)⟹f can be considered continuous at -77+x−3 is continuous on (−7,∞)1x2−4 is continuous on R∖{−2,2}⟹g(x) is continuous on [−7,−2)∪(−2,2)∪(2,∞)limx→−2+7+x−3x2−4=∞⟹−2 is an essential (second kind) discontinuity of g∀x≤9:f(x)=g(x)⟹−2 is an essential (second kind) discontinuity of flimx→27+x−3x2−4=limx→27+x−9(x−2)(x+2)(7+x+3)=limx→21(x+2)(7+x+3)=124⟹2 is a removable discontinuity of g∀x≤9:f(x)=g(x)⟹2 is a removable discontinuity of fx2+81 is continuous on R1x2−81 is continuous on R∖{−9,9}limx→9+f(x)=limx→9+x2+81x2−81=∞⟹9 is an essential (second kind) discontinuity of f⟹{f(x) is undefined for x<−7−2 is an essential (second kind) discontinuity of f2 is a removable discontinuity of f9 is an essential (second kind) discontinuity of f