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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 11 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 11
The value of $$\underset { x\rightarrow 0 }{ lim } \frac { { 27 }^{ x }-{ 9 }^{ x }{ -3 }^{ x }+1 }{ \sqrt { 5 } -\sqrt { 4+cos\quad x } } $$
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0%
$$\sqrt { 5 } { \left( log3 \right) }^{ 2 }$$
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$$8\sqrt { 5 } log3$$
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$$16\sqrt { 5 } log3$$
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$$8\sqrt { 5 } { \left( log3 \right) }^{ 2 }$$
$$\lim _ { x \rightarrow 0 } \dfrac { \sin 4 x } { \tan 7 x } =$$
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$$1$$
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$$\dfrac { 4 } { 7 }$$
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$$\dfrac { 7 } { 4 }$$
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$$0$$
The value of $$\underset { x\rightarrow 0 }{ lim } \frac { { 27 }^{ x }-{ 9 }^{ x }-{ 3 }^{ x }+1 }{ \sqrt { 5 } -\sqrt { 4+cosx } } $$ is
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0%
$$\sqrt { 5 } { \left( log3 \right) }^{ 2 }$$
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$$8\sqrt { 5 } log3$$
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$$16\sqrt { 5 } log3$$
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$$8\sqrt { 5 } { \left( log3 \right) }^{ 2 }$$
$$\displaystyle \lim _ { x \rightarrow 0 } \dfrac { | \cos ( \sin ( 3 x ) ) | - 1 } { x ^ { 2 } }$$ equals
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$$\dfrac { - 9 } { 2 }$$
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$$\dfrac { - 3 } { 2 }$$
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$$\dfrac { 3 } { 2 }$$
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$$\dfrac { 9 } { 2 }$$
$$\displaystyle\lim_{x\rightarrow \dfrac{\pi}{4}}\dfrac{\cos x-\sin x}{\left(\dfrac{\pi}{4}-x\right)(\cos x+\sin x)}=?$$
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$$2$$
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$$1$$
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$$0$$
0%
$$3$$
$$\displaystyle\lim_{x\rightarrow 0}\left\{\dfrac{log_e(1+x)}{x^2}+\dfrac{x-1}{x}\right\}$$ is equal to?
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$$\dfrac{1}{2}$$
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$$-\dfrac{1}{2}$$
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$$1$$
0%
None of these
Let [x] denote the greatest integer less than or equal to x. Then :
$$\underset { x\rightarrow 0 }{ lim } \dfrac { tan\left( \pi { sin }^{ 2 }x \right) +\left( \left| x \right| -sin \right) \left( x\left[ x \right] \right) ^{ 2 } }{ { x }^{ 2 } } :$$
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does not exist
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equal $$\pi $$
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equal 0
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equal $$\pi $$ + 1
$$\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } \frac { cotx-cosx }{ { (\pi -2x) }^{ 3 } } $$ equals :
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$$\frac { 1 }{ 8 } $$
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$$\frac { 1 }{ 4 } $$
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$$\frac { 1 }{ 24 } $$
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$$\frac { 1 }{ 16 } $$
$$\displaystyle\lim _{ n\rightarrow \infty }{ { n }^{ 2 }\left( { x }^{ \dfrac { 1 }{ n } }-{ x }^{ \dfrac { 1 }{ n+1 } } \right) ,x>0 } $$ is equal to
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0%
$$0$$
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$$e^{x}$$
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$$log_ex$$
0%
$$None\ of\ these$$
If $$\displaystyle \lim _{ x\rightarrow 0 }[1+ax+bx^{2}]^{(2/x)}=e^{3}$$, then
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$$a=3,\ b=0$$
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$$a=\dfrac{3}{2},\ b\neq1$$
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$$a=\dfrac{3}{2},\ b=R$$
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$$a=2,\ b=3$$
If [.] deotes the greatest integer function then
$$\begin{matrix} lim \\ x\rightarrow \pi /2 \end{matrix}\left[ \frac { x-\frac { \pi }{ 2 } }{ cosx } \right] $$ is equal to
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$$1$$
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$$-1$$
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$$2$$
0%
$$-2$$
If $$\begin{matrix} lim\quad \\ x\rightarrow 0 \end{matrix}\dfrac { x\left( 1+acosx \right) -bsinx }{ { x }^{ 3 } } =1$$ then value of a + b
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-4
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-6
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1
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None of these
$$\displaystyle x\xrightarrow { lim } a\left(\sin\frac{x-a}{2}\tan\frac{\pi x}{2a} \right) $$
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$$\displaystyle a/\pi$$
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$$-\displaystyle a/\pi$$
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$$\displaystyle \pi/a$$
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$$-\displaystyle \pi/a$$
Value of $$\underset { x\rightarrow 0 }{ lim } \dfrac { \sqrt [ 3 ]{ 1+\tan { x } } -\sqrt [ 3 ]{ 1-\tan { x } } }{ x } $$ is
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$$\dfrac { 1 }{ 2 } $$
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$$-\dfrac { 2 }{ 3 } $$
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$$\dfrac { 1 }{ 3 } $$
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$$0$$
Value of $$\underset { x\rightarrow \dfrac { \pi }{ 2 } }{ lim } \tan { x } .\ell nsin{ x }$$ is
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0
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$$\dfrac { 1 }{ 2 } $$
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$$\dfrac { 3 }{ 4 } $$
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None of these
The value of $$\displaystyle \lim_{x \rightarrow 0} (\sin x)^{\dfrac{1}{x}}+(1+x)^{(\sin x)})=0$$, where $$x > 0$$, is :
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$$0$$
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$$-1$$
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$$1$$
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$$2$$
The value of $$\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}$$ $$[\frac{x}{sinx}],$$ where [.] represents the greatest inter function , is
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$$1$$
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$$0$$
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$$-1$$`
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none of these
$$\underset { x\rightarrow 0 }{ lim } \left( \dfrac { 1+tanx }{ 1+sinx } \right) ^{ cosecx }$$ is equal to
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e
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$$\dfrac { 1 }{ e } $$
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1
0%
None of these
The value of $$\displaystyle \lim_{x\rightarrow 0}\left(\dfrac {1}{x^{2}}-\cot x\right)$$ equals
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$$1$$
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$$0$$
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$$\infty$$
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$$Does\ not\ exist$$
$$\underset { \theta \longrightarrow 0 }{ Lt } \dfrac { 3tan\theta -tan3\theta }{ { 2\theta }^{ 3 } } =$$
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-4
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1/4`
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3/4
0%
4
$$\underset { x\rightarrow 0 }{ Lt\quad } \frac { sec\quad x-1 }{ { \left( sec\quad x+\quad 1 \right) }^{ 2 } } =$$
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1/8
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11/4
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3.2
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2
$$\underset { x\rightarrow 0 }{ Lt } (1+sin\quad x)^{ cot\quad x }=$$
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e
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$$e^{ 2 }$$
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$$e^{ 3 }$$
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$$e^{ 4 }$$
For $$a > 1$$ then $$\displaystyle\lim_{x\rightarrow \infty}\dfrac{a^x-a^{-x}}{a^x+a^{-x}}=?$$
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$$1$$
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$$1/2$$
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$$1/3$$
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$$1/15$$
The value of $$\lim_{h\rightarrow 0}\left\{\dfrac {1}{h.(8+h)^{1/3}-\dfrac {1}{2h}}\right\}$$ equals
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0%
$$0$$
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$$\dfrac {-4}{3}$$
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$$\dfrac {-16}{3}$$
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$$\dfrac {-1}{48}$$
$$\underset { x\rightarrow 0 }{ lim } { \left( cosecx \right) }^{ \dfrac { 1 }{ logx } }$$ is equal to
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$$0$$
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$$1$$
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$$\dfrac { 1 }{ e } $$
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None of these
$$\underset { h\rightarrow 0 }{ lim } \quad \frac { (2+h)cos(2+h)-2cos2 }{ h } =$$
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cos2-2sin2
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cos2+2sin2
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sin2-2cos2
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sin2+2cos2
$$\underset { x\rightarrow 0 }{ lim } \dfrac { 1 }{ { e }^{ 2 } } tan\left( \dfrac { \pi }{ 4 } +x \right) ^{ 1/x }$$
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0
0%
1
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-1
0%
e
$$\lim_{x\rightarrow 1}\dfrac {1+\sin \pi\left(\dfrac {3x}{1+x^{2}}\right)}{1+\cos \pi x}$$ is equal to
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$$0$$
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$$1$$
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$$2$$
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$$4$$
$$lim_{x \to 0^-} \dfrac{x([x]+ \mid x \mid )sin[x]}{\mid x \mid}$$ is equal to
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$$-sin1$$
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0
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1
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$$sin1$$
The value of $$\displaystyle \lim _{ x\rightarrow \infty } (|x^{2}|+x)\log{(x\cot^{-1}{x})}$$ is :
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$$\dfrac{1}{3}$$
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$$-\dfrac{1}{3}$$
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$$\dfrac{2}{3}$$
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$$-\dfrac{2}{3}$$
$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } }{ \dfrac { \sin { x } }{ \cos ^{ -1 }{ \left[ \dfrac { 1 }{ 4 } \left( 3\sin { x } -\sin { 3x } \right) \right] } } } $$, where [.] denotes greatest integer function is :
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$$\dfrac{2}{\pi}$$
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$$1$$
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$$\dfrac{4}{\pi}$$
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$$does\ not\ exist$$
$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \left( \frac { 1 + \cos x } { 1 - \cos x } \right) ^ { \sec x } =$$
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0%
e
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$$e^2$$
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$$e^3$$
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$$e/4$$
$$Im _{ }{ \left( \dfrac { 1 }{ 1-\cos { \theta } +i\sin { \theta } } \right) } $$ is equal to
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$$\dfrac{1}{2}\tan\dfrac{\theta}{2}$$
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$$\dfrac{1}{2}\cot\dfrac{\theta}{2}$$
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$$-\dfrac{1}{2}\tan\dfrac{\theta}{2}$$`
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$$-\dfrac{1}{2}\cot\dfrac{\theta}{2}$$
The value of $$\lim_{x\rightarrow 0}\left(\dfrac {e^{x}+e^{-x}-2}{x^{2}}\right)^{1/x^{2}}$$ equals
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$$e^{1/2}$$
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$$e^{1/4}$$
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$$e^{1/3}$$
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$$e^{1/12}$$
$$\lim _ { x \rightarrow 0 } \frac { \sqrt [ 3 ] { 1 + \sin x } - \sqrt [ 3 ] { 1 - \sin x } } { x } =$$
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0%
0
0%
1
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3$$/ 2$$
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2$$/ 3$$
$$\displaystyle\underset{x\rightarrow \dfrac{\pi}{4}}{Lt}\dfrac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}=?$$
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$$\dfrac{1}{16\sqrt{2}}$$
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$$\dfrac{1}{32\sqrt{2}}$$
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$$\dfrac{1}{16}$$
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$$\dfrac{1}{8}$$
$$\lim _ { x \rightarrow 0 } \frac { \ln ( \sin 3 x ) } { \ln ( \sin x ) }$$ is equal to
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0%
0
0%
1
0%
2
0%
none of these
$$\underset { x\rightarrow \infty }{ Lt } { 5 }^{ x }sin\left( \cfrac { a }{ { 5 }^{ x } } \right) =$$
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0
0%
5
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log5
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None of these.
$$L\underset { x\rightarrow 0 }{ im } \frac { \sec { 4x-\sec { 2x } } }{ \sec { 3x-\sec { x } } }=$$
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0%
3/2
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2/3
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1/3
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3/4
the value of $$\underset { x\rightarrow 0 }{ lim } \frac { sin\alpha X-sin\beta X }{ { e }^{ \alpha X }-{ e }^{ \beta X } } $$ equals
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0%
0
0%
1
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-1
0%
$$\alpha -\beta $$
$$\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - 3\sin \left( {a + 2h} \right) + 3\sin \left( {a + h} \right) - \sin a}}{{{h^3}}}$$ is equal to
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$$\cos a$$
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$$ - \cos a$$
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$$\sin a$$
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$$\sin a\cos a$$
$$\lim _{ x\rightarrow -\infty }{ \cfrac { x^{ 4 }\sin { \cfrac { 1 }{ x } } +x^{ 2 } }{ 1+\left| x \right| ^{ 3 } } = } $$
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0%
1
0%
-1
0%
2
0%
-3
$$L\underset { x\rightarrow \infty }{ im } \left( \sin { \sqrt { x+1 } -\sin { \sqrt { x } } } \right) =$$
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0%
2
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-2
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0
0%
1
$$ \lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{x \sin 2 x}= $$
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1$$ / 2 $$
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3$$ / 2 $$
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3$$ / 4 $$
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1$$ / 4 $$
The value of $$\underset { x\rightarrow 0 }{ lim } \left( \left( sinx \right) ^{ 1/x }+\left( \dfrac { 1 }{ x } \right) ^{ sinx } \right) $$ equals
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0%
0
0%
1
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$$\infty $$
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-1
$$\displaystyle {Lt}_{x\rightarrow 0}\dfrac{cos5x cos3x}{x(sin5x sin3x)}$$ =
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$$-4$$
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$$4$$
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$$-\dfrac{1}{4}$$
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$$\dfrac{1}{4}$$
$$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right]$$ is equal to
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0%
1
0%
-1
0%
0
0%
None of these
The value of $$\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x } \right) }^{ 1/x }+{ \left( \dfrac { 1 }{ x } \right) }^{ \sin { x } } \right) } $$
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$$0$$
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$$1$$
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$$\infty$$
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$$-1$$
The value of $$\theta ,\quad is$$
$$\underset { o\rightarrow o }{ lim } \quad \frac { { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\quad \left( 1-{ cos }^{ 2 }\quad .....\left( { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\theta \right\} \right) \right) \right\} }{ sin\left( \frac { \pi (\sqrt { \theta +4 } -2 }{ \theta } \right) } $$
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$$\frac { \sqrt { 2 } }{ 4 } $$
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$$\sqrt { 2 } $$
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1
0%
2
$$\underset { x\rightarrow \infty }{ Lim } (sin\sqrt { x+1 } -sin\sqrt { x } )=$$
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0%
2
0%
-2
0%
0
0%
1
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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