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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity     Quiz 9 - MCQExams.com

Ltn3nr=2n+1nr2n2 is equal to :
  • ln23
  • ln32
  • ln23
  • ln32
The value of lim
  • \dfrac{-1}{4}
  • 1/2
  • 2
  • None of these
\displaystyle \lim_{I\rightarrow \left (\dfrac {\pi}{2}\right )} = \int_{0}^{t}\tan \theta \sqrt {\cos \theta} ln (\cos \theta) d\theta is equal to
  • -4
  • 4
  • -2
  • Does not exists
If f ' (0) = 0 and f(x) is a differentiable and increasing function,then lim x \rightarrow 0  \frac {x.f ' (x^2)}{f ' (x)}
  • is always equal to zero
  • may not exist as left hand limit may not exist
  • may not exist as left hand limit may not exist
  • right hand limit is always zero
Consider f(x)=\lim _{ n\rightarrow \infty  }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } }  }{ { x }^{ n }+\sin { { x }^{ n } }  }  } for x>0,x\neq 1,f(1)=0 then
  • f is continuous at x=1
  • f has a discontinuity at x=1
  • f has an infinite or oscillatory discontinuity at x=1
  • f has a removal type of discontinuity at x=1
If \sum _{ r=1 }^{ k }{ \cos ^{ -1 }{ \beta  }  } =\cfrac { k\pi  }{ 2 } for any k\ge 1 and A=\sum _{ r=1 }^{ k }{ { \left( { \beta  }_{ r } \right)  }^{ r } } , then \lim _{ x\leftarrow A }{ \cfrac { { \left( 1+x \right)  }^{ 1/3 }-{ \left( 1-2x \right)  }^{ 1/4 } }{ x+{ x }^{ 2 } }  } is equal to
  • 0
  • \cfrac{1}{2}
  • \cfrac { \pi }{ 2 }
  • \cfrac { 5 }{ 6 }
\displaystyle \lim_{x \rightarrow 0} \frac{ae^x + b cos x + c. e^{-x}}{sin^2 x} = 4 then b =
  • 2
  • 4
  • -2
  • -4
\lim_{x\rightarrow \dfrac{\pi}{6}} \dfrac{2 sin^{2} x + sin x - 1}{2 sin^{2} x - 3 sin x + 1}
  • 6
  • -6
  • -3
  • 3
Let f:R \rightarrow  (0,1) be a continuous function.. Then, which of the following function(s) has (have) the value zero at some point in the interval (0,1)?
  • e ^ { x } - \int _ { 0 } ^ { 1 } f ( t ) \sin t d t
  • f ( x ) + \int _ { 0 } ^ { 1 } f ( t ) \sin t d t
  • x - \int _ { 0 } ^ { \frac { \pi } { 2 } - x } f ( t ) \cos t d t
  • x ^ { 3 } - f ( x )
\mathop {\lim }\limits_{x \to {a^ + }} {{\left\{ x \right\}\sin \left( {x - a} \right)} \over {{{\left( {x - a} \right)}^2}}} 

is equal to (where {.} denotes the fraction
part of x and a \in N

  • 0
  • 1
  • does not exist
  • none of thes
If f\left( x \right) = \left\{ \begin{array}{l}\frac{{1 - \left| x \right|}}{{1 + x}},{\rm{ }}x \ne  - 1\\1,{\rm{          }}x =  - 1{\rm{     }}\end{array} \right.   then f\left( {\left[ {2x} \right]} \right), where \left[ {} \right] represents the greatest integer function , is 
  • discontinuous at x = - 1
  • continuous at x = 0
  • continuous at x = \frac{1}{2}
  • continuous at x = 1
If f : R \rightarrow R is defined by
f(x) = \left \{\begin{matrix} \dfrac{x + 2}{x^2 + 3x + 2} & if & x \in R - \{-1, -1\} \\ -1 & if & x = -2 \\ 0 & if  &x = -1\end{matrix} \right. then f(x) continuous on the set 
  • R
  • R - \{-2\}
  • R - \{-1, -2\}
  • R-\{-1\}
If l=\lim\limits_{n\to 3}\dfrac{x^2-9}{\sqrt{x^2+7}-4} and m=\lim\limits_{n\to -3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}, then
  • l\ne m
  • l=2m
  • l=-m
  • l=m
Letf(\theta) = \dfrac{1}{tan^{9}\theta} {(1+tan\theta)^{10}+(2+tan\theta)^{10}+....+(20+tan\theta)^{10}}-20tan\theta. The left hand limit of f(\theta) as \theta \rightarrow \dfrac{\pi}{2} is:
  • 1900
  • 2000
  • 2100
  • 2200
Consider A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix}, then the value of \lim_{n \rightarrow \infty} \dfrac{A^{n}}{n} (where \theta \in R) is equal to 
  • 10
  • zero matrix
  • symmetric matrix
  • 4
If \displaystyle\lim_ { x\rightarrow \lambda  } { \left( 2-\dfrac { \lambda  }{ x }  \right)  }^{ \lambda tan\left( \dfrac { \pi x }{ 2\lambda  }  \right)  }=\frac { 1 }{ e } , then \lambda is equal to-
  • -\pi
  • \pi
  • \dfrac{\pi }{2}
  • -\dfrac{2}{\pi }
\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right) has the value equal to
  • \dfrac {a-b}{3}
  • 0
  • \dfrac {(a^{2}-b^{2})}{6a^{2}b^{2}}
  • \dfrac {a^{2}-b^{2}}{3a^{2}b^{2}}
\displaystyle \lim _{ x\rightarrow 2 }{ \frac { \sqrt [ 3 ]{ 60+{ x }^{ 2 } } -4 }{ \sin { \left( x-2 \right)  }  }  } 
  • \dfrac {1}{4}
  • 0
  • \dfrac {1}{12}
  • Does\ not\ exist
if\left( x \right)= greatest integer \le x, then \underset { x\longrightarrow 2 }{ lim }  { \left( -1 \right)  }^{ \left[ x \right]  } is  equal to- is equal to-
  • 0
  • -1
  • +1
  • none of these
The value of \displaystyle \lim _{ x\rightarrow 0 }{ \csc^{ 4 }{ x } \int _{ 0 }^{ { x }^{ 2 } }{ \frac { ln\left( 1+4t \right)  }{ { t }^{ 2 }+1 }  } dt }  is 
  • 1
  • 2
  • 3
  • 4
The value of \displaystyle\lim_{n\rightarrow \infty}n(n\{ln (n)-ln (n+1)\}+1) is?
  • e
  • \dfrac{1}{e}
  • \dfrac{1}{2}
  • \dfrac{1}{4}
\displaystyle \lim_{x\rightarrow \infty}{x^{2}\sin\left(\log_{e}\sqrt{\cos\dfrac{\pi}{x}}\right)}
  • 0
  • -\dfrac{\pi^{2}}{2}
  • -\dfrac{\pi^{2}}{4}
  • -\dfrac{\pi^{2}}{8}
 \underset { x\rightarrow \cfrac { \pi  }{ 2 }  }{ lim } \cfrac { cot \,  x-cos\, x }{ \left( \pi -{ 2x } \right)^ 3 } equals
  • \cfrac {1} {24}
  • \cfrac {1} {16}
  • \cfrac {1} {8}
  • \cfrac {1} {4}
\underset { x\rightarrow \frac { \pi  }{ 2 }  }{ lim } \frac { (1-sinx)({ 8x }^{ 2 }-{ \pi  }^{ 3 })cosx }{ { (\pi -2x) }^{ 4 } }
  • -\cfrac { { \pi }^{ 2 } }{ 16 }
  • \cfrac { { 3\pi }^{ 2 } }{ 16 }
  • \cfrac { { \pi }^{ 2 } }{ 16 } \quad
  • -\cfrac { 3{ \pi }^{ 2 } }{ 16 } \quad
 \underset { x\rightarrow 0 }{ lim } \left[ { x }^{ 2 }cosec\quad \left( { x }^{ 2 } \right)^0 \right]  is equal to :
  • \cfrac { \pi} {180}
  • \cfrac { \pi} {90}
  • {0} 
  • \cfrac {90} { \pi }
The value of \lim_{x \rightarrow -1} \dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} is given by 
  • \dfrac{1}{\sqrt{2}\sqrt{\pi}}
  • \dfrac{1}{\sqrt{\sqrt{2\pi}}}
  • 1
  • 0
If \lim_{x \rightarrow 0}\dfrac{a \sin x-bx+cx^{2}+x^{3}}{2x^{2} \log(1+x)-2x^{3}+x^{4}} exists and is finite, then the value of a,b,c are respectively 
  • 0,6,6
  • 6,0,6
  • 6,6,0
  • 0,0,6
 \underset { x\rightarrow a }{ lim } \cfrac { sin\quad x-sin\quad a }{ \sqrt [ 3 ]{ x } -\sqrt [ 3 ]{ a }  } 
  • \sqrt [ 3 ]{ a\quad cos\quad a }
  • 2\sqrt [ 3 ]{ a }
  • { 3a }^{ { 2 /}{ 3 } }cos\quad a
  • \sqrt [ 2 ]{ a\quad cos\quad a }
The value of \displaystyle \lim_{x\rightarrow 0}\dfrac {1-\cos^{3}x}{x\sin x\cos x} is
  • \dfrac {2}{5}
  • \dfrac {3}{5}
  • \dfrac {3}{2}
  • \dfrac {3}{4}
Integrate:
 lim_{x\rightarrow 0}\dfrac{(1-\cos{2x})^{2}}{2x\tan{x}-x\tan{2x}}
  • 2
  • \dfrac{-1}{2}
  • -2
  • \dfrac{1}{2}
\displaystyle \lim_{x\rightarrow 0^{+}}{(\csc x)^{1/\log x}}=
  • e
  • e^{-1}
  • e^{2}
  • 1
The value of \lim_{x \rightarrow 0} \left(\dfrac{\tan x}{x}\right)^{1/x^{3}} is-
  • 0
  • \infty
  • e^{1/4}
  • Does\ not\ exist
The value of \underset { x\rightarrow \frac { x }{ 2 }  }{ lim } \frac { log\sin { x }  }{ { \left( \frac { \pi  }{ 2 } -x \right)  }^{ 2 } } is 
  • 0
  • \frac{1}{2}
  • -\frac{1}{2}
  • -2
\lim_{n\rightarrow \infty}\dfrac{1}{n^{2}}\left[\sin^{3}\dfrac{\pi}{4n}+2\sin^{3}\dfrac{2\pi}{4n}+3\sin^{3}\dfrac{3\pi}{4n}+....+n\sin^{3}\dfrac{n\pi}{4n}\right]=
  • \dfrac{\sqrt{2}}{9\pi^{2}}\left(52-15\pi\right)
  • \dfrac{\sqrt{2}}{9\pi^{2}}\left(52+15\pi\right)
  • \dfrac{\sqrt{2}}{9\pi}\left(52-17\pi\right)
  • \dfrac{\sqrt{2}}{9\pi^{2}}\left(52+17\pi\right)
If \alpha \quad and \beta are the roots of the equation  {ax}^{2}+bx+c=0 , then 
  \underset { x\rightarrow \cfrac { \pi  }{ 2 }  }{ lim } \cfrac { tan\left[ \left( \alpha +\beta  \right) x \right]  }{ sin\left[ \left( \alpha \beta  \right) x \right]  }  is equal to :
  • \cfrac {c} {b}
  • - \cfrac {b} {c}
  • \cfrac {a} {b}
  • - \cfrac {a} {b}
\begin{matrix} lim \\ n\rightarrow \infty  \end{matrix}\int _{ 0 }^{ 1 }{ \frac { { nx }^{ n-1 } }{ 1+{ x }^{ 2 } }  } dx=
  • 0
  • 1
  • 2
  • \frac { 1 }{ 2 }
Arrange the following limits in the ascending order :
(1)  \lim _ { x \rightarrow \infty } \left( \dfrac { 1 + x } { 2 + x } \right) ^ { x + 2 }

(2)  \lim _ { x \rightarrow 0 } ( 1 + 2 x ) ^ { 3 / x }

(3)  \lim _ { \theta \rightarrow 0 } \dfrac { \sin \theta } { 2 \theta }

(4)  \lim _ { x \rightarrow 0 } \dfrac { \log _ { e } ( 1 + x ) } { x }
  • 1,2,3,4
  • 1,3,4,2
  • 1,4,3,2
  • 3,4,1,2
\mathop{\lim}\limits_{x \to 0} \left(\dfrac{3+x}{3-x}\right)^{\dfrac{x+1}{x}} is equal to 
  • e^{2/3}
  • e^{1/3}
  • e^3
  • e^2
If \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 1, then
  • a = \frac{5}{2}
  • b = \frac{{ - 5}}{2}
  • a + b = 4
  • a + b = -4
\lim_{x\rightarrow 0 }(\frac{p^{\frac{1}{x}}+q^{\frac{1}{x}}+r^{\frac{1}{x}}+s^{\frac{1}{x}}}{4})3x where p,q,r,s> 0 is equal to
  • pqrs
  • (pqrs)^{3}
  • (pqrs)\frac{3}{2}
  • (pqrs)\frac{3}{4}
\lim- {x\to 0} \dfrac{1- cos(1 - cos4x)}{x^4} is equal to : 
  • 4
  • 16
  • 32
  • None of these
The value of \displaystyle\lim_{x\to 0} |x|^{sinx} equals 
  • 0
  • -1
  • 1
  • does not exist
If \displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \left( \sin { nx }  \right) \left[ (a-n)nx-tanx \right]  }{ { x }^{ 2 } }  } =0, then the value of a
  • \dfrac { 1 }{ n }
  • n-\dfrac { 1 }{ n }
  • n+\dfrac{1}{n}
  • None\ of\ these
\lim _ { x \rightarrow 0 } \frac { 1 - \cos x \cos 2 x \cos 3 x } { \sin ^ { 2 } 2 x } =
  • \frac { 3 } { 2 }
  • \frac { 5 } { 2 }
  • \frac { 7 } { 4 }
  • \frac { 9 } { 2 }
The value of lim_{x \to 0} (\dfrac{1}{x^2} - cotx) equals 
  • 1
  • 0
  • \infty
  • Does not exist
\displaystyle \lim _{ x-\infty  }{ sgn\left( \cot{\dfrac { { \pi x }^{ 2019 } }{ { x }^{ 2019 }+7 }}  \right)  }
  • Equals -1
  • Equals 1
  • equals 0
  • Does not exit
\displaystyle\lim_{x \to \pi/2} (sec x +tan x) is equal to 
  • 1
  • -1
  • \dfrac{1}{2}
  • 0
\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)=
  • 0
  • -\dfrac{1}{2}
  • \dfrac{1}{2}
  • 1
\underset { x\rightarrow 0 }{ lim } \dfrac { x\tan { 2x } -2\tan { 2x }  }{ { \left( 1-cos2x \right)  } } equals:
  • \dfrac{1}{4}
  • 1
  • \dfrac{1}{2}
  • -\dfrac{1}{2}
\displaystyle\lim_{x\to \pi/2} \dfrac{sinx-(sinx)^{sin x}}{1-sin x + In sin x} is equal to
  • 4
  • 2
  • 1
  • none of these
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers