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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity     Quiz 10 - MCQExams.com

limnΣnr=1πnsin(πrn) is equal to
  • 1
  • 2
  • 3
  • 4
Evaluate : limx0(exn(3x1)(3x1)xsinxexnx) is equal to 
  • 1en3
  • e n 3
  • 3
  • 13
Ltx0sinxx+x3/6x5
  • 1120
  • 1110
  • 1100
  • none of these
limx0(cosα)x(sinα)xcos2α(x4),α(0,π2) is equal to
  • cos4α.log(cosα)sin4α.log(sinα)
  • sin4α.log(cosα)cos4α.log(sinα)
  • sin4α.log(cosα)+cos4α.log(sinα)
  • None of the above
limx03sin(x9)sin(x9)x3=q
  • 0
  • 4(π200)3
  • π200
  • π100
Ltx0tanxxx2tanx equals:
  • 1
  • 1/2
  • 1/3
  • None of these
evaluatelimx0xx0cost2dtx36x
  • 3
  • 1
  • 0
  • 1
limx0{tan(π4+x)}1x=
  • e
  • e2
  • e1
  • e2
limx(x+12x+1)x2 equals?
  • 0
  • e
  • 1
If g(x)=x[x]forx>2thenLimx2+g(x)g(2)x2
  • -1
  • 0
  • 1/2
  • 1
limx2tan1xπ equals eL then L is equal to
  • 2π
  • 2π
  • π2
  • 1
limxx/2[1tan(x2)][1sinx][1+tan(x2)][π2x]3 is
  • 18
  • 0
  • 132
limθ04θ(tanθ2θtanθ)(1cos2θ)2 is
  • 1/2
  • 1/2
  • 1
  • 2
the value of limxx55x is-
  • 0
  • 1
  • e5
  • e5
If k  is an integer such that limn[(coskπ4)2(coskπ6)2]=0 then :
  • k is divisible neither by 4 nor by 8.
  • k must be divisible by 12 but not necessarily by 24.
  • k must be divisible by 24.
  • either k is divisible by 24 or k is neither divisible by 4 nor by 6.
limx5sin2(x5)tan(x5)(x225)(x5)=
  • 1/10
  • 1
  • 0
  • 6
The value of limx0cosec4xx20In(1+4t)1+t2dt is
  • 1
  • 2
  • 3
  • 4
limnn2(x1/nx1/(n+n)),x>0, is equal to 
  • 0
  • ex
  • logex
  • none of these
limx0ln(sin3x)ln(sinx) is equal to
  • 0
  • 1
  • 2
  • Non existent
limxn/2cotxcosx(π2x)3 equals
  • 124
  • 116
  • 18
  • 14
limx2360+x24sin(x2) equals 
  • 14
  • 0
  • 112
  • Does not exist
limx0sinxsin(π3+x)sin(π3x)x=
  • 34
  • 14
  • 43
  • 0
limx02(3sin(π6+x)cos(π6+x))x3(3cosxsinx)
  • 1/3
  • 2/3
  • 4/3
  • 4/3
limxπ/422(cosx+sinx)21sin2x is equal to 
  • 1
  • 22
  • 423
  • does not exist
Let f be a differentiable function such that f(x)=734f(x)x,(x>0) and f(1)4.
Then limx0+xf(1x):
  • Exists and equals 4
  • Does not exist
  • Exist and equals
  • Exists and equals 47
If Limx0(1+a3)+8e1/x1+(1b3)+e1/x=2 then
  • a=1,b=2
  • a=1,b=31/3
  • a=2,b=31/3
  • none of these
limxπ2tan2x(2sin2x+3sinx+4sin2x+6sinx+2) is equal to
  • 34
  • 16
  • 112
  • 512
limx01cosxxlog(1+x)=
  • 1
  • 0
  • 1
  • 12
Evaluate limx0sin[cosx]1+[cosx] ([.] denotes the greatest integer function)
  • 1
  • 1
  • 0
  • does not exist
limx0x(esinx1)1cosx
  • 12
  • 2
  • 0
  • 1
P=limxo+(1+tan2x)1/2x=____
  • P=1/4
  • P=1/2
  • P=2
  • P=1
limx0(11/x+21/x+31/x+.....n1/xn)nx, n ϵ N is equal to
  • n!
  • 1
  • 1n!
  • 0
The values of limn4n5+23n2+15n4+22n3+1 is?
  • 1
  • 0
  • 1
limx0xtan2x2xtanx(1cos2x)2 equal to 
  • 14
  • 1
  • 12
  • Noneofthese
Let f(x)=limn2x2nsin1x+x1+x2x then find 
  • limxxf(x)
  • limx1f(x)
  • limx0f(x)
  • limxf(x)
If limx0(cosx+a3sin(b6x))1x=e512 then value of ab2 is equal to
  • 512
  • 512
  • 8
  • none of these
if(x)=[x3]+[x4]forxϵRthenlimx3f(x)=
  • -2
  • -1
  • 0
  • 2
limx0(cosx)1/2(cosx)1/3sin2x is $$
  • 1/6
  • 1/12
  • 2/3
  • 1/3
The value of limxysin2xsin2yx2y2
  • 0
  • 1
  • sinyy
  • sin2y2y
limx0xtan2x2xtanx(1cos2x)2 is equal to?
  • 2
  • 2
  • 1/2
  • 1/2
limx0(cosx+asinbx)1x is equal to 
  • ea
  • eab
  • eb
  • ea/b
limx0x(1+acosx)bsinxx3=1 then
  • a=5/2
  • a=3/2,b=1/2
  • a=3/2,b=5/2
  • a=5/2,b=3/2
If limx0[1+xln(1+b2)]1/x=2bsin2θ,b> s m f θ(π,π], then the value of θ is?
  • ±π4
  • ±π3
  • ±π6
  • ±π2
 limxsin(πcos2x)x2  is equal to:
  • π
  • π
  • π2
  • 1
The value of limθ0+sinθsinθ is equal to?
  • 0
  • 1
  • 1
  • 4
Let f : (π2,π2)R , f(x)={limn(tanx)2n+x2sin2x+(tanx)2n;x01;x=0,nN . Which of the following holds good ?
  • f(π4)=f(π+4)
  • f(π4)=f(π+4)
  • f(π4)=f(π+4)
  • f(0+)=f(0)=f(0)
limx0(1sin2x1sinh2x)=?
  • 2/3
  • 0
  • 1/3
  • 2/3
 Consider  limx0ax+bex+sinx+1axbsinx=(  is a finite number)
  • =0, if a=2,b=1
  • =5,ifa=1,b=1
  • =1, if a=b=1
  • None of these
limx0sin(πcos2x)x2 is equal to?
  • π
  • π
  • π/2
  • 1
If f(x) is odd linear polynomial with f(1)=1, then limx02f(tanx)2f(sinx)x2f(sinx) is 
  • 1
  • ln2
  • 12ln2
  • cos2
0:0:1


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