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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity     Quiz 2 - MCQExams.com

If x is very large, then 2x1+x is
  • close to 0
  • arbitrarily large
  • lie between 2 and 3
  • close to 2
What is limx0cosxπx equal to?
  • 0
  • π
  • 1π
  • 1
limx0(1+ax)bx is equal to
  • eab
  • ea+b
  • non-existent
  • none of these
limx0 ((1+x)2ex)4sinx is:
  • e2
  • e4
  • e8
  • e8
If a sequence <an> is such that a1,an+1=2+3an1+2an and limnan exists, then an is equal to
  • cos36o
  • 2cos36o
  • sin18o
  • 2sin36o
limx 0sin7xsin3x equals
  • 73
  • 103
  • 143
  • 13
limx0(1+sinx)cosx is equal to 
  • 0
  • e
  • 1
  • 1e
Evaluate the following limit :
limx01cos2xx2
  • 0
  • 1
  • 2
  • none of these
Evaluate limn1+2+3+...+nn2
  • 12
  • 1
  • 32
  • 122
Which of the following statement is not correct
  • limxc[f(x)+g(x)]=limxcf(x)+limxcg(x)
  • limxc[f(x)g(x)]=limxcf(x)limxcg(x)
  • limxc[f(x).g(x)]=limxcf(x).limxcg(x)
  • limxcf(x)g(x)=limxcf(x)limxcg(x)
What is the value of limx0sinxtan3x
  • 14
  • 13
  • 12
  • 1
If f(x)={x5;x14x29;1<x23x2+4;x>2
Then f(2+)f(2)=
  • 0
  • 2
  • 9
  • 4
 If f(x)={2x+b(x<α)x+d(xα)is such that
limxαf(x)=L, then L=
  • 2db
  • bd
  • d+b
  • b2d

limx1(1x)tan(πx2)=
  • π
  • 2π
  • π2
  • 2π
lf f(x)=x,x<0;f(x)=0,x=0; f(x)=x2;x>0, then limx0f(x) is equal to
  • Does not exist
  • 0
  • 1
  • 1
f(x)=2x+1,a=1,l=3 and ϵ=0.001, then δ>0 satisfying 0<|xa|<δ such that |f(x)l|<ϵ, is

  • 0.0005
  • 0.005
  • 0.001
  • 0.0001
 Evaluate: limh0(1h38+h12h)
  • 112
  • 43
  • 163
  • 148
 lf limxa+f(x)=L, then for each ϵ>0, there exists δ>0 so that
  • 0<|xa|<δ|f(x)L|ϵ
  • 0<|xa|<δ|f(x)L|<ϵ
  • a<x<a+δf(x)L<ϵ
  • aδ<x<a|f(x)L|<ϵ
limx01cosxxlog(1+x)=
  • 1
  • 0
  • 1
  • 12
limx0(1ex)sinxx2+x3=
  • 1
  • 0
  • 1
  • 2
lf f(x)=2x3,a=2,l=1 and ϵ=0.001 then δ>0 satisfying0<|xa|<δ,  |f(x)l|<ϵ, is:
  • 0.005
  • 0.0005
  • 0.001
  • 0.0001
limx0exesinx2(xsinx)=
  • 12
  • 12
  • 1
  • 32

limx0xtan2x2xtanx(1cos2x)2=
  • 2
  • 2
  • 12
  • 12
lf limx0(cos4x+acos2x+bx4) is finite then the value of a,b respectively are
  • 54
  • 5, 4
  • 4, 3
  • 4, 5

limx01cos3xxsin2x=
  • 12
  • 32
  • 34
  • 14

limxπ2(π2x)secxcosecx=
  • 1
  • 0
  • -1
  • 1π
Solve : limx0sinxsin(π3+x)sin(π3x)x
  • 34
  • 14
  • 43
  • 0
limx0(1cos2x)sin5xx2sin3x=
  • 103
  • 310
  • 65
  • 56
The value of limαβ[sin2αsin2βα2β2] is:
  • 0
  • 1
  • sinββ
  • sin2β2β
Evaluate: limx0sec4xsec2xsec3xsecx
  • 32
  • 23
  • 13
  • 34

limxπ2cosecxcotxx=
  • 1
  • 2π
  • 12
  • 15
limxπ4secx.tan(4xπ)sin(4xπ)=
  • 2
  • 12
  • 2
  • 12
limxπ63sinx3cosx6xπ=
  • 3
  • 13
  • 3
  • 13
limx0tanx0x=
  • 0
  • 1
  • π180
  • π

limx03sinxsin3xx3=
  • 0
  • 1
  • (π180)3
  • 4.(π180)3
Solve:
limx03tanxtan3x2x3
  • 14
  • 34
  • 4
  • 4

limx0axbxx=
  • log(ab)
  • log(ba)
  • log(ab)
  • logba
limx0(sinxxx)(sin1x) is:
  • does not exist
  • is equal to 0
  • is equal to 1
  • exists and different from 0 and 1

limxπ4cosxsinx(π4x)(cosx+sinx)=
  • 2
  • 1
  • 0
  • 3

limxπ21sinθcosθ(π2θ)=
  • 1
  • 1
  • 12
  • 12

limxπ21sinx(π2x)2=
  • 13
  • 14
  • 16
  • 18
limxx+sinxx+cosx=
  • 1
  • 1
  • 13
  • 0
f(x)={3x2sin2x2,x<0x2+2x+C13x2,x0,x130,x=13
limx0+f(x)=
  • C4
  • 2C
  • C3
  • C
Let f(x)={x21,0<x<22x+3,2x<3, 
The quadratic equation whose roots are limx2f(x) and limx2+f(x) is
  • x210x+21=0
  • x26x+9=0
  • x214x+49=0
  • x2+6x+9=0
The value of ee upto four decimal places is?
  • 1.6484
  • 1.5237
  • 1.2589
  • 1.9190
If f(x)=x2+6xsinx , then limx0f(x)=
  • 2
  • 4
  • 6
  • 8

limx0log(1+ax)log(1+bx)x=
  • a+b
  • ab
  • (a+b)
  • ab
Let f be a continuous function on [1,3]. lf f takes only rational values for all x and f(2)=10 then f(1.5) is equal to
  • 8
  • f(1)+f(3)3
  • 20
  • 10
Evaluate: limx0  sin3x2cos(2x2x)
  • 0
  • 1
  • 4
  • 6
limx03sin(2x2)x2=A
then the value of A is
  • 2
  • 4
  • 6
  • 8
0:0:1


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