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

$$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$$ is equal to
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
Evaluate : $$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right)  }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x }  }{ { e }^{ x\ell nx } }  \right)  } $$ is equal to 
  • $$\dfrac{1}{e}\ell n3$$
  • $$e\ \ell n\ 3$$
  • $$3$$
  • $$\dfrac{1}{3}$$
$$\underset{x \rightarrow 0}{Lt} \dfrac{sin x - x + x^3 / 6}{x^5}$$ = 
  • $$\dfrac{1}{120}$$
  • $$\dfrac{1}{110}$$
  • $$\dfrac{1}{100}$$
  • none of these
$$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$$ is equal to
  • $$\cos^{4}\alpha.\log(\cos\alpha)-\sin^{4}\alpha.\log(\sin\alpha)$$
  • $$\sin^{4}\alpha.\log(\cos\alpha)-\cos^{4}\alpha.\log(\sin\alpha)$$
  • $$\sin^{4}\alpha.\log(\cos\alpha)+\cos^{4}\alpha.\log(\sin\alpha)$$
  • $$None\ of\ the \ above$$
$$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { 3\sin { \left( { x }^{ 9} \right) -\sin { \left( { x }^{ 9 } \right)  }  }  }{ { x }^{ 3 } }  } =q$$
  • 0
  • $$4\left(\dfrac{\pi}{200}\right)^{3}$$
  • $$\dfrac{\pi}{200}$$
  • $$\dfrac{\pi}{100}$$
$$\underset { x\rightarrow 0 }{ Lt } \cfrac {tanx-x}{x^2tanx}$$ equals:
  • 1
  • 1/2
  • 1/3
  • None of these
evaluate$$ \underset { x\rightarrow 0 }{ lim } \frac { x-\int _{ 0 }^{ x }{ { cost }^{ 2 }dt }  }{ { x }^{ 3 }-6x } $$
  • $$3$$
  • $$-1$$
  • $$0$$
  • $$1$$
$$\lim _ { x \rightarrow 0 } \left\{ \tan \left( \frac { \pi } { 4 } + x \right) \right\} ^ { \frac { 1 } { x } } =$$
  • $$e$$
  • $$e ^ { 2 }$$
  • $$e ^ { - 1 }$$
  • $$e ^ { - 2 }$$
$$\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x+1}{2x+1}\right)^{x^2}$$ equals?
  • $$0$$
  • e
  • $$1$$
  • $$\infty$$
If $$g(x)=\frac { x }{ \left[ x \right]  } for\quad x>2\quad then\quad \underset { x\rightarrow { 2 }^{ + } }{ Lim } \frac { g\left( x \right) -g\left( 2 \right)  }{ x-2 } $$
  • -1
  • 0
  • 1/2
  • 1
$$\underset{x \rightarrow \infty}{lim} \dfrac{2 \tan^{-1} x}{\pi}$$ equals $$e^L$$ then $$L$$ is equal to
  • $$\dfrac{2}{\pi}$$
  • $$-\dfrac{2}{\pi}$$
  • $$-\dfrac{\pi}{2}$$
  • $$1$$
$$\displaystyle \lim _{ x\rightarrow x/2 } \dfrac { \left[ 1-\tan { \left( \dfrac { x }{ 2 }  \right)  }  \right] \left[ 1-\sin { x }  \right]  }{ \left[ 1+\tan { \left( \dfrac { x }{ 2 }  \right)  }  \right] \left[ \pi -2x \right] ^{ 3 } } $$ is
  • $$\dfrac {1}{8}$$
  • $$0$$
  • $$\dfrac {1}{32}$$
  • $$\infty$$
$$\displaystyle \lim _{ \theta \rightarrow 0 }{ \frac { 4\theta \left( \tan { \theta -2\theta \tan { \theta  }  }  \right)  }{ { \left( 1-\cos { 2\theta  }  \right)  }^{ 2 } }  } $$ is
  • $$1/\sqrt {2}$$
  • $$1/2$$
  • $$1$$
  • $$2$$
the value of $$\underset { x\rightarrow \infty  }{ lim } \frac { { x }^{ 5 } }{ { 5 }^{ x } } $$ is-
  • 0
  • 1
  • $${ e }^{ 5 }$$
  • $${ e }^{ -5 }$$
If $$k$$  is an integer such that $$\lim_{n \rightarrow \infty} \left[\left(\cos \dfrac{k\pi}{4}\right)^{2}-\left(\cos \dfrac{k\pi}{6}\right)^{2}\right]=0$$ 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 $$6$$.
$$\lim _ { x \rightarrow 5 } \frac { \sin ^ { 2 } ( x - 5 ) \tan ( x - 5 ) } { \left( x ^ { 2 } - 25 \right) ( x - 5 ) } =$$
  • 1$$/ 10$$
  • $$1$$
  • $$0$$
  • $$-6$$
The value of $$\underset{x \rightarrow 0}{lim} \cos \,ec^4 \,x \displaystyle \overset{x^2}{\underset{0}{\int}} \dfrac{In(1 + 4t)}{1 + t^2}dt$$ is
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
$$\lim_{n \rightarrow \infty}n^{2}\left(x^{1/n}-x^{1/\left(n+n\right)}\right),x>0$$, is equal to 
  • $$0$$
  • $$e^{x}$$
  • $$\log_{e}x$$
  • $$none\ of\ these$$
$$\lim_{x\rightarrow 0}\frac{ln(sin 3x)}{ln(sin x)}$$ is equal to
  • 0
  • 1
  • 2
  • Non existent
$$\underset{x \rightarrow n/2}{lim} \dfrac{\cot x - \cos x}{(\pi - 2x)^3}$$ equals
  • $$\dfrac{1}{24}$$
  • $$\dfrac{1}{16}$$
  • $$\dfrac{1}{8}$$
  • $$\dfrac{1}{4}$$
$$\underset{x \rightarrow 2}{lim} \dfrac{\sqrt[3]{60 + x^2} - 4}{\sin (x - 2)}$$ equals 
  • $$\dfrac{1}{4}$$
  • $$0$$
  • $$\dfrac{1}{12}$$
  • Does not exist
$$\lim _ { x \rightarrow 0 } \frac { \sin x \sin \left( \frac { \pi } { 3 } + x \right) \sin \left( \frac { \pi } { 3 } - x \right) } { x } =$$
  • $$\frac { 3 } { 4 }$$
  • $$\frac { 1 } { 4 }$$
  • $$\frac { 4 } { 3 }$$
  • $$0$$
$$\lim_{x\rightarrow 0}\dfrac{2\left(\sqrt{3}\sin\left(\dfrac{\pi}{6}+x\right)-\cos\left(\dfrac{\pi}{6}+x\right)\right)}{x\sqrt{3}\left(\sqrt{3}\cos x-\sin x\right)}$$
  • $$-1/3$$
  • $$2/3$$
  • $$4/3$$
  • $$-4/3$$
$$\lim_{x\rightarrow \pi/4}\dfrac{2\sqrt{2}-\left(\cos x+\sin x\right)^{2}}{1-\sin 2x}$$ is equal to 
  • 1
  • $$2\sqrt{2}$$
  • $$\dfrac{4\sqrt{2}}{3}$$
  • $$does\ not\ exist$$
Let $$f$$ be a differentiable function such that $$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$$ and $$f(1) \neq 4$$.
Then $$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right) $$:
  • Exists and equals 4
  • Does not exist
  • Exist and equals
  • Exists and equals $$\dfrac{4}{7}$$
If $$\underset { x\rightarrow 0 }{ Lim } \dfrac { \left( 1+{ a }^{ 3 } \right) +8{ e }^{ 1/x } }{ 1+\left( 1-{ b }^{ 3 } \right) +{ e }^{ 1/x } } =2$$ then
  • $$a=1,b=2$$
  • $$a=1,b={-3}^{1/3}$$
  • $$a=2,b={3}^{1/3}$$
  • none of these
$$lim_{x\to \dfrac{\pi}{2}} tan^2x(\sqrt{2sin^2x + 3 sin x +4} - \sqrt{sin^2x + 6 sin x+2})$$ is equal to
  • $$\dfrac{3}{4}$$
  • $$\dfrac{1}{6}$$
  • $$\dfrac{1}{12}$$
  • $$\dfrac{5}{12}$$
$$\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x \log ( 1 + x ) } =$$
  • $$1$$
  • $$0$$
  • $$-1$$
  • $$\frac { 1 } { 2 }$$
Evaluate $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \sin { \left[ \cos { x }  \right]  }  }{ 1+\left[ \cos { x }  \right]  }  } $$ ($$[.]$$ denotes the greatest integer function)
  • $$-1$$
  • $$1$$
  • $$0$$
  • $$does\ not\ exist$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ \sin { x }  }-1 \right)  }{ 1-\cos { x }  }  } $$
  • $$\dfrac {1}{2}$$
  • $$2$$
  • $$0$$
  • $$1$$
$$P= \lim_{x \rightarrow o^{+}} (1+ \tan^{2} \sqrt{x})^{1/2x}=$$____
  • $$P=1/4$$
  • $$P=1/2$$
  • $$P=2$$
  • $$P=1$$
$$\displaystyle \lim _{ x\rightarrow 0 } \left(\dfrac{1^{1/x}+2^{1/x}+3^{1/x}+.....n^{1/x}}{n}\right)^{nx} ,\ n\ \epsilon \ N $$ is equal to
  • $$n!$$
  • $$1$$
  • $$\dfrac{1}{n!}$$
  • $$0$$
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$$
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers