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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 \displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt[2]{n^3+1}} is?
  • 1
  • 0
  • -1
  • \infty
\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { x\tan { 2x } -2x\tan { x }  }{ \left( 1-\cos { 2x }  \right) ^{ 2 } }  } equal to 
  • \dfrac{1}{4}
  • 1
  • \dfrac{1}{2}
  • None of these
Let f(x)=\displaystyle\lim _{ n\rightarrow \infty  }{ \dfrac { { 2x }^{ 2n }\sin { \frac { 1 }{ x } +x }  }{ 1+{ x }^{ 2x } }  } then find 
  • \displaystyle\lim _{ x\rightarrow \infty }{ xf\left( x \right) }
  • \displaystyle\lim _{ x\rightarrow 1 }{ f\left( x \right) }
  • \displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right) }
  • \displaystyle\lim _{ x\rightarrow -\infty }{ f\left( x \right) }
If \lim _{x \rightarrow 0}\left(\cos x+a^{3} \sin \left(b^{6} x\right)\right)^{\frac{1}{x}}=e^{512}  then value of ab^2 is equal to
  • -512
  • 512
  • 8
  • none of these
if\left( x \right) =\left[ x-3 \right] +\left[ x-4 \right] \quad for\quad x\epsilon R\quad then\quad \underset { x\rightarrow 3 }{ lim } f\left( x \right) =
  • -2
  • -1
  • 0
  • 2
\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { { \left( \cos { x }  \right)  }^{ 1/2 }-{ \left( \cos { x }  \right)  }^{ 1/3 } }{ \sin ^{ 2 }{ x }  }  } is $$
  • 1/6
  • -1/12
  • 2/3
  • 1/3
The value of \begin{matrix} lim \\ x\rightarrow y \end{matrix}\dfrac { { sin }^{ 2 }x-sin^{ 2 }y }{ { x }^{ 2 }-{ y }^{ 2 } }
  • 0
  • 1
  • \dfrac { siny }{ y }
  • \dfrac { sin2y }{ 2y }
\displaystyle\lim_{x\rightarrow 0}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2} is equal to?
  • 2
  • -2
  • 1/2
  • -1/2
\underset { x\rightarrow 0 }{ lim } (\cos  x+a\sin  b{ x) }^{ \frac { 1 }{ x }  } is equal to 
  • e^a
  • e^{ab}
  • e^b
  • e^{a/b}
\displaystyle \lim_{x\rightarrow 0}{\dfrac{x(1+a\cos x)-b\sin x}{x^{3}}}=1 then
  • a=-5/2
  • a=-3/2,b=-1/2
  • a=-3/2,b=-5/2
  • a=-5/2,b=-3/2
If \displaystyle\lim_{x\rightarrow 0}[1+x ln (1+b^2)]^{1/x}=2b\sin^2\theta, b > s m f \theta\in (-\pi, \pi], then the value of \theta is?
  • \pm \dfrac{\pi}{4}
  • \pm \dfrac{\pi}{3}
  • \pm \dfrac{\pi}{6}
  • \pm \dfrac{\pi}{2}
 \lim _ { x \rightarrow \infty } \dfrac { \sin \left( \pi \cos ^ { -2 } x \right) } { x ^ { 2 } }  is equal to:
  • - \pi
  • \pi
  • \dfrac { \pi } { 2 }
  • 1
The value of \displaystyle\lim_{\theta \rightarrow 0^+}\dfrac{\sin\sqrt{\theta}}{\sqrt{\sin\theta}} is equal to?
  • 0
  • 1
  • -1
  • 4
Let f : \left ( \dfrac{\pi}{2}, \dfrac{\pi}{2} \right )\rightarrow R , f(x) = \left \{\begin{matrix} \lim_{n\rightarrow \infty }\dfrac{(tanx)^{2n} + x^2}{sin^2x + (tanx)^{2n}}; & x \neq 0 \\ 1; & x = 0 \end{matrix} \right., n \in N . Which of the following holds good ?
  • f \left( -\dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)
  • f \left( -\dfrac{\pi^-}{4} \right) = f \left (-\dfrac{\pi^+}{4} \right)
  • f \left( \dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)
  • f(0^+) = f(0) = f(0^)
\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?
  • 2_{\displaystyle /3}
  • 0
  • 1_{\displaystyle /3}
  • -2_{\displaystyle /3}
 Consider  \lim _ { x \rightarrow 0 } \dfrac { a x + b e ^ { - x } + \sin x + 1 } { a x - b \sin x } = \ell ( \ell  is a finite number)
  • \ell = 0 , \text { if } a = - 2 , b = - 1
  • \ell = 5 , i f a = 1 , b = - 1
  • \ell = 1 , \text { if } a = b = - 1
  • None of these
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin(\pi \cos^2x)}{x^2} is equal to?
  • -\pi
  • \pi
  • \pi/2
  • 1
If f(x) is odd linear polynomial with f(1)=1, then \underset{x \to 0} {\lim} \dfrac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)} is 
  • 1
  • \ln 2
  • \dfrac{1}{2}\ln 2
  • \cos 2
0:0:2


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