CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity     Quiz 6 - MCQExams.com

$${ x }_{ n }={ \left( 1-\cfrac { 1 }{ 3 }  \right)  }^{ 2 }{ \left( 1-\cfrac { 1 }{ 6 }  \right)  }^{ 2 }{ \left( 1-\cfrac { 1 }{ 10 }  \right)  }^{ 2 }...{ \left( 1-\cfrac { 1 }{ \cfrac { n(n+1) }{ 2 }  }  \right)  }^{ 2 },n\ge 2$$. Then the value of $$\displaystyle\lim _{ n\rightarrow \infty  }{ { x }_{ n } } $$
  • $$1/3$$
  • $$1/9$$
  • $$1/81$$
  • $$0$$ (zero)
$$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}$$ is equal to
  • $$1$$
  • $$1/2$$
  • $$1/3$$
  • $$0$$
The value of k which makes $$f(x)=\left\{\begin{matrix} \sin\dfrac{1}{x}, x\neq 0\\ k, x=0\end{matrix}\right.$$ continuous at $$x=0$$ is?
  • $$8$$
  • $$1$$
  • $$-1$$
  • None
If $$f(x) = 2x^9 - 5x^8 + 7x^6 - 15x^4 + 5x + 7$$, then $$\underset{x \rightarrow 0}{\lim} \dfrac{f (1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is 
  • $$-\dfrac{35}{3}$$
  • $$\dfrac{35}{3}$$
  • $$\dfrac{37}{3}$$
  • $$-\dfrac{37}{3}$$
The value of $$\displaystyle \lim _{ x\rightarrow 0 } \dfrac{1+\sin{x}-\cos{x}+\log{(1-x)}}{x^{3}}$$, is
  • $$-1$$
  • $$1/2$$
  • $$-1/2$$
  • $$1$$
If $$f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$$, then $$\underset{a \rightarrow 0}{\lim} \dfrac{f(1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is 
  • $$-\dfrac{53}{3}$$
  • $$\dfrac{53}{3}$$
  • $$-\dfrac{55}{3}$$
  • $$\dfrac{55}{3}$$
Let $$f(x) = \underset{0}{\overset{x}{\int}} |2t - 3| dt$$, then $$f$$ is 
  • continuous at $$x = \dfrac{3}{2}$$
  • continuous at $$x = 3$$
  • differentiable at $$x = \dfrac{3}{2}$$
  • differentiable at $$x = 0$$
The value of $$\lim_{x\rightarrow o}\dfrac{\sqrt{\dfrac{1}{2}(1-cos 2 x)}}{x}$$
  • 1
  • -1
  • 0
  • none of these
solve the limit 
$$\mathop {\lim }\limits_{x \to 3} \dfrac{2}{{x - 3}}$$ 
  • $$2$$
  • $$3$$
  • $$4$$
  • Does not exist
The value of $$\underset { x\longrightarrow \infty  }{ Lim } \dfrac{d}{dx}\overset { \sqrt { 3 }  }{ \underset { -\sqrt { 3 }  }{ \int }  } \dfrac{r^3}{(r+1)(r-1)}$$dr,is
  • $$0$$
  • $$1$$
  • $$\dfrac{1}{2}$$
  • non existent
$$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to
  • $$1$$
  • $$0$$
  • $$2$$
  • $$1/2$$
$$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} = $$
  • $$\dfrac{-1}{2}$$
  • $$\dfrac{1}{2}$$
  • $$2$$
  • $$-2$$
$$\mathop {\lim }\limits_{x \to 0} \,\dfrac{1}{x}\,{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to
  • $$1$$
  • $$0$$
  • $$2$$
  • $$\dfrac{1}{2}$$
Solve

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}$$
  • $$-\dfrac{5}{3}$$
  • $$\dfrac{5}{3}$$
  • $$\dfrac{7}{3}$$
  • None of these
$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{2{x^2} + x - 3}} = $$
  • $$\dfrac{1}{{10}}$$
  • $$\dfrac{{ - 1}}{{10}}$$
  • $$\dfrac{2}{5}$$
  • $$ - \dfrac{2}{5}$$
If $$ \mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) } $$ is non-
zero finite $$ ( \mathrm { a } > 0 ) , $$ then-
  • L = 2 , b = 1 , c = 1
  • $$

    L = \frac { 1 } { 2 } , b = 1 , c = 1

    $$
  • L = 4 , b = - 1 , c = - 1
  • $$

    L = \frac { 1 } { 4 } , b = - 1 , c = - 1

    $$
If   $${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$$, where $$ r= 1, 2, 3, ....n$$, then $$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right)$$ is equal to 
  • $$\cos \alpha + i\sin \alpha $$
  • $$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$$
  • $${e^{i\alpha /2}}$$
  • $$\sqrt {{e^{i\alpha }}} $$
Solve:
$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=$$
  • $$\dfrac{4}{3}(\sqrt{2}+1)$$
  • $$\dfrac{4}{3}(\sqrt{2}-1)$$
  • $$\dfrac{3}{4}(\sqrt{2}-1)$$
  • $$\dfrac{3}{4}(\sqrt{2}-2)$$
If $$f(x)$$ is the integral of $$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$$. Find $$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right)  } $$, where $$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$$
  • $$\dfrac{1}{2}$$
  • $$1$$
  • $$\dfrac{1}{3}$$
  • $$2$$
$$\lim_{x\to \infty} \dfrac{\sqrt{x^2 + sin^2x}}{x+cosx}$$ equals
  • $$1$$
  • $$0$$
  • $$\infty$$
  • does not exist
Find the value of limit $$\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 }  }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 }  } = } $$.
  • $$0$$
  • $$3$$
  • $$-3$$
  • $$1$$
$$\lim\limits_{x\rightarrow0}\dfrac{x\tan 2x-2x\tan x}{\left(1-\cos 2x\right)^{2}}=$$
  • $$2$$
  • $$\dfrac{1}{2}$$
  • $$-2$$
  • $$-\dfrac{1}{2}$$
$$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}=$$__________.
  • $$\cos\sqrt{x}$$
  • $$\dfrac{1}{2\sin \sqrt{x}}$$
  • $$\dfrac{\cos \sqrt{x}}{2\sqrt{x}}$$
  • $$\sin\sqrt{x}$$
The value of $$\displaystyle lim_{x\to 0} \dfrac{cos (sin x) - cos x}{x^4} $$ is equal to
  • $$1/5$$
  • $$1/6$$
  • $$1/4$$
  • $$1/2$$
The value of $$\lim_{x \rightarrow 1} \sec \dfrac{\pi}{2x} \log x$$ is-
  • $$ \pi /2$$
  • $$ 2 /\pi$$
  • $$-\pi/2$$
  • $$-2/ \pi$$
The value of $$\displaystyle \lim_{\theta \rightarrow 0^{+}} \dfrac {\sin \sqrt {\theta}}{\sqrt {\sin  \theta}}$$ is equal to
  • $$0$$
  • $$1$$
  • $$-1$$
  • $$4$$
For $$x>y$$, $$\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]}$$ is :
  • 0
  • -1
  • 1
  • 2
$$\underset{x\rightarrow 0}{lim} \dfrac{\sqrt{a^2 -ax+x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} -\sqrt{a-x}}$$ is equal to (a > 0)
  • $$\sqrt{a}$$
  • $$-\sqrt{a}$$
  • $$a \sqrt{a}$$
  • $$-a\sqrt{a}$$
The function $$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$$ is
  • discontinuous at only one point
  • discontinuous exactly at two points
  • discontinuous exactly at three points
  • None of these
$$\displaystyle\lim _{ x\rightarrow \dfrac { x }{ 2 }  }{ \dfrac { \cot { x } -\cos { x }  }{ \left( \pi -2x \right) ^{ 3 } }  } $$ is equal to 
  • $$ \dfrac{1}{24} $$
  • $$ \dfrac{1}{16} $$
  • $$ \dfrac{1}{8} $$
  • $$ \dfrac{1}{4} $$
$$\lim _ { x \rightarrow 1 } \{ 1 - x + [ x + 1 ] + [ 1 - x ] \} , \text { where } [ x ]$$ denotes greatest integer function is
  • $$0$$
  • $$1$$
  • $$-1$$
  • $$2$$
$$ \underset { x\rightarrow 0 }{ lim } \cfrac { 1+cos\left( \pi x \right)  }{ \left( 1-x \right)^ 2 }   $$ is equal to :
  • $$ \cfrac { \pi} {2} $$
  • $$ - \cfrac { \pi ^2} {2} $$
  • $$\cfrac { \pi ^2} {2} $$
  • $$ \cfrac { \pi} {3} $$
Evaluate: $$\underset{x\rightarrow 0}{lim} \dfrac{e^{1/x} - 1}{e^{1/x }+ 1}$$
  • $$0$$
  • $$1$$
  • $$-1$$
  • does not exist
If $$\underset { x\rightarrow { 0 } }{ lim } \dfrac { \left( ax+b \right) -\sqrt { 4+\sin x }  }{ \tan\quad x } =\dfrac { 27 }{ 4 } ~where ~a,b\in R$$ then the value of 
  • $$a = 2~ and~ b = 7$$
  • $$a = -2~and~ b = 7 $$
  • $$a = 7~ and~ b = 2 $$
  • $$a=7~ and ~b = -2$$
Evaluate: $$\underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } } $$ 
  • $$\dfrac { 1 }{ 4 } $$
  • $$1$$
  • $$\dfrac { 1 }{ 2 } $$
  • $$-\dfrac { 1 }{ 2 } $$
$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$$ equals:
  • $$\frac { 1 } { 24 }$$
  • $$\frac { 1 } { 16 }$$
  • 0
  • $$\frac { 1 } { 4 }$$
Evaluate: $$\underset { x\rightarrow { 0 } }{ lim } \dfrac { x\tan 2x-2x \tan\ x\quad  }{ (1-\cos2x)^{ 2 } } $$
  • $$\dfrac { 1 }{ 4 } $$
  • $$1$$
  • $$\dfrac { 1 }{ 2 } $$
  • $$-\dfrac { 1 }{ 2 } $$
$$\underset {x\rightarrow0}{ lim } \dfrac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^2}$$ equals 
  • $$ \dfrac{1}{4}$$
  • 1
  • $$ \dfrac{1}{2}$$
  • $$-\dfrac{1}{2}$$
Solve:
$$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} =$$
  • $$\sqrt{2}.2^{\sqrt{2}}$$
  • $$2^{\sqrt{2}- 1}$$
  • $$2^{\sqrt{2} - \dfrac{1}{2}}$$
  • $$2^{\sqrt{2}}$$
Let $$f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad  when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}$$. If the function $$f(x)$$ be continous at $$x=0$$, then $$k=$$
  • $$0$$
  • $$1$$
  • $$2$$
  • $$-2$$
If $$2f(\sin x)+\sqrt {2}f(\cos x)=\tan x,\ (x> 0)$$, then $$\displaystyle \lim _{ x\rightarrow 1 }{ \sqrt { 1-x } f\left( x \right) = }$$ 
  • $$\sqrt {2}$$
  • $$\dfrac {1}{\sqrt {2}}$$
  • $$-\sqrt {2}$$
  • $$-\dfrac {1}{\sqrt {2}}$$
Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow a}\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$$.
  • $$2\sqrt{a}$$
  • $$2{a}$$
  • $$2{a^{\frac 13}}$$
  • None of these
$$Lt_{x\to 0} \dfrac{sin x - x+\dfrac{x^3}{6}}{x^5}=$$_________
  • $$\dfrac{1}{120}$$
  • $$\dfrac{-1}{120}$$
  • $$0$$
  • $$\dfrac{1}{6}$$
The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } $$ where $$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$$, is
  • $$2$$
  • $$\dfrac {1}{6}$$
  • $$\dfrac {2}{3}$$
  • $$-\dfrac {1}{3}$$
$$\lim _{ x\rightarrow \infty  }{ \frac { 1 }{ x } \int _{ 0 }^{ x }{ \left( \sqrt { { t }^{ 2 }+5t } -t \right) dt }  }$$ 
  • $$0$$
  • $$1$$
  • $$\dfrac{5}{2}$$
  • $$5$$
The value of $$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b }  }{ { x }^{ 2 }-{ a }^{ 2 } }  } \left( a>b \right) $$ is
  • $$\dfrac {1}{4a}$$
  • $$\dfrac {1}{a\sqrt {a-b}}$$
  • $$\dfrac {1}{2a\sqrt {a-b}}$$
  • $$\dfrac {1}{4a\sqrt {a-b}}$$
$$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } }  \right]  }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right)  }{ { x }^{ 3 }-{ x }^{ -1 } }  \right] =$$
  • $$\frac { 1 }{ 3 } $$
  • 3
  • $$\frac { 1 }{ 2 } $$
  • $$\frac { 3 }{ 2 } $$
The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x }{ 5 }  } \left[ \frac { x }{ 2 }  \right] $$ (where $$[.]$$ denotes the greatest integer function) is
  • $$\dfrac{2}{5}$$
  • $$-\dfrac{2}{5}$$
  • $$0$$
  • $$\infty$$
The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}$$ is
  • $$10/3$$
  • $$3/10$$
  • $$6/5$$
  • $$5/6$$
Let  $$U_{ { n } }=\dfrac { n! }{ (n+2)! } $$  where  $$n \in N .$$  If  $$S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }$$  then  $$\lim _ { n \rightarrow \infty } \mathrm { S } _ { n }$$  equals :
  • $$2$$
  • $$1$$
  • $$1/2$$
  • $$1/3$$
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers