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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: \displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)
  • \displaystyle \frac{1}{12}
  • -\displaystyle \frac{4}{3}
  • -\displaystyle \frac{16}{3}
  • -\displaystyle \frac{1}{48}
 lf \displaystyle \lim_{x\rightarrow a^+}f(x)=L, then for each \epsilon>0, there exists \delta>0 so that
  • 0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon
  • 0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon
  • a < x < a+\delta\Rightarrow f(x) - L <\epsilon
  • a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon
\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=
  • 1
  • 0
  • -1
  • \displaystyle \dfrac{1}{2}
\displaystyle \lim_{x\rightarrow 0}\frac{(1-e^{x})\sin x}{x^{2}+x^{3}}=
  • -1
  • 0
  • 1
  • 2
lf f(x)=2x-3,a=2,l=1 and \epsilon =0.001 then \delta>0 satisfying 0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon, is:
  • 0.005
  • 0.0005
  • 0.001
  • 0.0001
\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=
  • -\dfrac{1}{2}
  • \dfrac{1}{2}
  • 1
  • \dfrac{3}{2}

\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}=
  • 2
  • -2
  • \displaystyle \frac{1}{2}
  • -\displaystyle \frac{1}{2}
lf \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle  \frac { \cos  4x+a\cos  2x+b }{ x^{ 4 } }  \right) is finite then the value of a,b respectively are
  • 5\, -4
  • -5,\ -4
  • -4,\ 3
  • 4,\ 5

\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=
  • \displaystyle \frac{1}{2}
  • \displaystyle \frac{3}{2}
  • \displaystyle \frac{3}{4}
  • \displaystyle \frac{1}{4}

\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}=
  • 1
  • 0
  • -1
  • \displaystyle \dfrac{1}{\pi}
Solve : \displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}
  • \displaystyle \frac{3}{4}
  • \displaystyle \frac{1}{4}
  • \displaystyle \frac{4}{3}
  • 0
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=
  • \dfrac{10}{3}
  • \dfrac{3}{10}
  • \dfrac{6}{5}
  • \dfrac{5}{6}
The value of \displaystyle \lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin ^{ 2 }{ \alpha  } -\sin ^{ 2 }{ \beta  }  }{ { \alpha  }^{ 2 }-{ \beta  }^{ 2 } }  \right]  } is:
  • 0
  • 1
  • \displaystyle \frac { \sin { \beta  }  }{ \beta  }
  • \displaystyle \frac { \sin { 2\beta  }  }{ 2\beta  }
Evaluate: \displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}
  • \displaystyle \frac{3}{2}
  • \displaystyle \frac{2}{3}
  • \displaystyle \frac{1}{3}
  • \displaystyle \frac{3}{4}

\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}=
  • 1
  • \displaystyle \frac{2}{\pi}
  • \displaystyle \frac{1}{2}
  • \displaystyle \frac{1}{5}
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}=
  • \sqrt{2}
  • \displaystyle \frac{1}{\sqrt{2}}
  • -\sqrt{2}
  • \displaystyle \frac{-1}{\sqrt{2}}
\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=
  • \sqrt{3}
  • \dfrac{1}{\sqrt{3}}
  • -\sqrt{3}
  • -\dfrac{1}{\sqrt{3}}
\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=
  • 0
  • 1
  • \displaystyle \frac{\pi}{180}
  • \pi

\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}=
  • 0
  • 1
  • (\displaystyle \frac{\pi}{180})^{3}
  • 4.(\displaystyle \frac{\pi}{180})^{3}
Solve:
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}
  • \dfrac{1}{4}
  • \dfrac{3}{4}
  • 4
  • -4

\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}=
  • \displaystyle \log(\frac{a}{b})
  • \displaystyle \log(\frac{b}{a})
  • \log(ab)
  • {\log_{b}}{a}
\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x}) is:
  • does not exist
  • is equal to 0
  • is equal to 1
  • exists and different from 0 and 1

\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}=
  • 2
  • 1
  • 0
  • 3

\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=
  • 1
  • -1
  • -\displaystyle \frac{1}{2}
  • \displaystyle \frac{1}{2}

\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}=
  • \displaystyle \frac{1}{3}
  • \displaystyle \frac{1}{4}
  • \displaystyle \frac{1}{6}
  • \displaystyle \frac{1}{8}
\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=
  • 1
  • -1
  • \dfrac{1}{3}
  • 0
\displaystyle f(x)=\left\{ \begin{matrix} \dfrac { 3 }{ { x }^{ 2 } } \sin { 2{ x }^{ 2 } } ,\; x<0 \\ \dfrac { { x }^{ 2 }+2x+C }{ 1-3{ x }^{ 2 } } ,\; x\ge 0,\; x\neq \frac { 1 }{ \sqrt { 3 }  }  \\ 0,\; x=\frac { 1 }{ \sqrt { 3 }  }  \end{matrix} \right.
\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=
  • \frac{C}{4}
  • 2C
  • \frac{C}{3}
  • C
Let f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases}, 
The quadratic equation whose roots are \displaystyle \lim_{x\rightarrow 2^{-}}f(x) and \displaystyle \lim_{x\rightarrow 2^{+}}f(x) is
  • x^{2}-10x+21=0
  • x^{2}-6x+9=0
  • x^{2}-14x+49=0
  • x^{2}+6x+9=0
The value of \sqrt{e}e upto four decimal places is?
  • 1.6484
  • 1.5237
  • 1.2589
  • 1.9190
If f(x) = \displaystyle \frac {x^2+6x}{\sin x} , then \displaystyle \lim_{x\rightarrow 0^{-}}f(x)=
  • 2
  • 4
  • 6
  • 8

\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}=
  • a+b
  • a-b
  • -(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
  • \displaystyle \frac{f(1)+f(3)}{3}
  • 20
  • 10
Evaluate: \displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}
  • 0
  • -1
  • 4
  • -6
\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { 3\sin(2{ x }^{ 2 }) }{ { x }^{ 2 } }  } =A
then the value of A is
  • 2
  • 4
  • 6
  • 8
0:0:2


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