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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 1 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 1
Evaluate $$\underset{x \rightarrow 3}\lim \sqrt[4] {x^3}$$ using the properties of limits.
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$$28^{1/4}$$
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$$25^{1/4}$$
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$$27^{1/4}$$
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$$26^{1/4}$$
Explanation
$$\displaystyle \lim _{ x\rightarrow 3 }{ \sqrt [ 4 ]{ { x }^{ 3 } } } \\={ (3^3) }^{ { 1 }/{ 4 } }={ (27) }^{ { 1 }/{ 4 } }$$
$$If {A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}}, \,i = 1,2,3,.....n$$ and $${a_1}< {a_2}< {a_3}....< {a_{n,}} \, then$$
$$\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}......{A_n}} \right), 1 \le m \le n$$
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is equal to $${\left( { - 1} \right)^m}$$
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is equal to $${\left( { - 1} \right)^{m + 1}}$$
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is equal to $${\left( { - 1} \right)^{n-m}}$$
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is equal to $${\left( { - 1} \right)^{n-m-1}}$$
$$\displaystyle \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\frac{\pi }{2} - x} \right)}^3}}}$$
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$$\dfrac{-1}{2}$$
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$$\dfrac{1}{2}$$
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$$2$$
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$$1$$
$$f(x)= x\sin\dfrac{1}{x} , \ for x\neq 0$$
$$= 0,\ for x=0$$
Then.
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$$f'(0^+) exit\ but \ f'(0^-)$$ does not exit
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$$f'(0^+) \ and \ f'(0^-)$$ do not exit
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$$f'(0^+) = f'(0^-)$$
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none of these
$$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } $$ is equal to
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0%
$$0$$
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$$1$$
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$$-1$$
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$$\dfrac{1}{2}$$
If $$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right)\, = 4$$,then
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$$a=1,b=4$$
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$$a=1,b=-4$$
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$$a=2,b=-3$$
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$$a=2,b=3$$
$$\displaystyle\lim_{n\rightarrow \infty}\left(\tan\theta +\dfrac{1}{2}\tan \dfrac{\theta}{2}+\dfrac{1}{2^2}\tan \dfrac{\theta}{2^2}+...+\dfrac{1}{2^n}\tan\dfrac{\theta}{2^n}\right)$$ equals?
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$$\dfrac{1}{\theta}$$
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$$\dfrac{1}{\theta}-2\cot 2\theta$$
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$$2\cot 2\theta$$
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None of these
Let p= $$\lim_{x\rightarrow 0+}(1+tan^{2}\sqrt{x})^{\frac{1}{2x}}$$ then log p is equal to :
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1
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$$\frac{1}{2}$$
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$$\frac{1}{4}$$
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2
$$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 4 } }{ { \left( \sin { 2x } \right) }^{ \sec ^{ 2 }{ 2x } } }$$ is equal to
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$$-\dfrac {1}{2}$$
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$$\dfrac {1}{2}$$
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$$e^{-\dfrac {1}{2}}$$
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$$e^{\dfrac {1}{2}}$$
$$\displaystyle \lim _{ \theta \rightarrow \pi /2 }{ \dfrac { 1-\sin \theta }{ (\pi /2-\theta )\cos { \theta } } } $$ is equal to
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$$1$$
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$$-1$$
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$$1/2$$
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$$-1/2$$
If $$\underset {x \rightarrow \infty}{lim} (1+\frac {a}{x}-\frac {4}{x^{2}})^{2x} =e^{3}$$ , then 'a' is equal to :
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2
0%
$$\frac {3}{2}$$
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$$\frac {2}{3}$$
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$$\frac {1}{2}$$
The value of $$lim_{x\to 0} \dfrac{sin(\pi cos^2 x)}{x^2}$$ equals
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$$-\pi$$
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$$\pi$$
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$$\dfrac{\pi}{2}$$
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$$2 \pi$$
$$\underset { x\rightarrow 1 }{ lim } \frac { xtan(x-[x]) }{ x-1 } $$ is:
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1
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0
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-1
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does not exist
The value of $$lim_{\theta \to \dfrac{\pi}{2}} (sec \theta - tan \theta)$$ equals
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$$0$$
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$$1$$
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$$2$$
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$$\infty$$
The value of $$\lim _{ x\rightarrow 0 }{ \dfrac { 1-\cos { ^{ 3 }x } }{ x\sin { x\cos { x } } } }$$ is
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$$\dfrac{2}{5}$$
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$$\dfrac{3}{5}$$
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$$\dfrac{3}{2}$$
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$$\dfrac{3}{4}$$
The value of $$\displaystyle\lim _ { x \rightarrow \infty } a ^ { x } \sin \left( \frac { b } { a ^ { x } } \right)$$ where $$a > 1$$ is
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$$b \log a$$
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$$a \log b$$
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$$b$$
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$$a$$
The value of $$\displaystyle \lim _{ x\rightarrow 3 }{ \dfrac { \left( { x }^{ 3 }+27 \right) \log _{ e }{ \left( x-2 \right) } }{ { x }^{ 2 }-9 } } $$ is
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$$9$$
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$$18$$
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$$27$$
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$$\dfrac {1}{3}$$
The value of $$\displaystyle \lim _{ x\rightarrow \frac { x }{ 4 } }{ \dfrac { 4\sqrt { 2 } -{ \left( \cos { x } +\sin { x } \right) }^{ 5 } }{ 1-\sin { 2x } } } $$ is
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$$0$$
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$$\sqrt {2}$$
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$$5\sqrt {2}$$
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$$3$$
$$\displaystyle \lim _ { x \rightarrow 0 } \left( \left[ \dfrac { - 5 \sin x } { x } \right] + \left[ \dfrac { 6 \sin x } { x } \right] \right)$$ (where $$[ .]$$ denotes greatest integer function) is equal to
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$$0$$
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$$-12$$
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$$1$$
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$$2$$
If $$\underset { x\rightarrow 0 }{ lim } \dfrac { { x }^{ 3 } }{ \sqrt { a+x } (bx-sinx) } =1,$$ a > 0, then a + b is equal to
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36
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37
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38
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40
$$\lim _ { x \rightarrow 0 } \dfrac { \sin x ^ { 5 } } { \sin ^ { 4 } x } =$$
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$$0$$
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$$1$$
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$$-1$$
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$$\infty$$
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}=?$$
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$$10/3$$
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$$3/10$$
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$$6/5$$
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$$56$$
$$\underset { x\rightarrow 0 }{ lim } \dfrac { 1-\cos { 2x } }{ \cos { 2x } -\cos { 8x } } $$ is equal to
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0%
$$-1/15$$
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$$1/10$$
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$$1/15$$
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$$15$$
$$\lim _ { x \rightarrow - 1 } \dfrac { \cos 2 - \cos 2 x } { x ^ { 2 } - | x | }$$ is equal to :
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$$0$$
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$$\cos 2$$
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$$2 \sin 2$$
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None
$$\lim _{ x\rightarrow 0 }{ \cfrac { x.{ 10 }^{ x }-x }{ 1-cosx } = } $$
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$$\log { 10 } $$
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$$2\log { 10 } $$
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$$3\log { 10 } $$
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$$4\log { 10 } $$
$$\lim _{ x\rightarrow 0 }{ \frac { \sqrt [ 3 ]{ 1+\sin { x } } -\sqrt [ 3 ]{ 1-\sin { x } } }{ x } } =$$
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$$0$$
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$$1$$
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$$\frac { 2 }{ 3 } $$
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$$\frac { 3 }{ 2 } $$
$$\underset { x\rightarrow 0 }{ lim } \frac { sin({ 6x }^{ 2 }) }{ Incos({ 2x }^{ 2 }-x) } =$$
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12
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-12
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6
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-6
$$\lim _ { x \rightarrow 1 } \left( \log _ { 3 } 3 x \right) ^ { \log _ { x } 3 } =?$$
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$$e ^ { - 1 }$$
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$$e$$
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$$-1$$
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$$1$$
Let $$f ( \beta ) = \lim _ { \alpha \rightarrow \beta } \dfrac { \sin ^ { 2 } \alpha - \sin ^ { 2 } \beta } { \alpha ^ { 2 } - \beta ^ { 2 } },$$ then $$f \left( \dfrac { \pi } { 4 } \right)$$ is greater than-
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$$\lim _ { x \rightarrow 0 } \dfrac { 1 - \cos ^ { 3 } x } { x \sin 2 x }$$
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$$\lim _ { x \rightarrow \pi / 2 } \dfrac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$$
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$$\lim _ { x \rightarrow \infty } ( \cos \sqrt { x + 1 } - \cos \sqrt { x } )$$
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$$\lim _ { x \rightarrow a } \dfrac { \sqrt { a + 2 x } - \sqrt { 3 x } } { \sqrt { 3 a + x } - 2 \sqrt { x } }$$ where $$a > 0$$
$$\underset { x\rightarrow -\infty }{ lim } \frac { ({ 3x }^{ 4 }+{ 2x }^{ 2 })sin(\frac { 1 }{ x } )+{ |x| }^{ 3 }+5 }{ { |x }|^{ 3 }+{ |x| }^{ 2 }+|x|+1 } =$$
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2
0%
1
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-2
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-3
The value f $$\lim_{x\rightarrow \pi/4}\dfrac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}=$$
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$$-\dfrac{1}{4}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{2}$$
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$$None\ of\ these$$
If $$L = \underset{x \rightarrow 0}{lim} \dfrac{a \, sin \, x - sin \, 2x}{tan^3 x}$$ is finite, then the value of L is :
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1
0%
2
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3
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-1
If $$\displaystyle \lim_{x\rightarrow 0}\dfrac {ae^{-x}-b\cos x-\dfrac {1}{2}cx}{x\cos x}=2$$ then the value of $$a+b+c$$ is-
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$$4$$
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$$-4$$
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$$2$$
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$$-2$$
$$\displaystyle \lim_{x\rightarrow 0} \dfrac {\sin 2x + 3x}{2x + \sin 3x}$$ is equal to
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$$1$$
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$$\dfrac {1}{5}$$
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$$2$$
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Does not exist
If $$f(x)$$ is continuous and $$f\left(\dfrac {9}{2}\right)=\dfrac {2}{9}$$, then $$\displaystyle\lim _{ x\rightarrow 0 }f\left(\frac {1-\cos 3x}{x^2}\right)$$ is equal to :
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$$\dfrac {9}{2}$$
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$$\dfrac {2}{9}$$
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0
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$$\dfrac {8}{9}$$
Explanation
$$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { 1-cos3x }{ x^{ 2 } } }$$
$$ =\lim _{ x\rightarrow 0 }{ \dfrac { 2\sin ^{ 2 }{ (\frac { 3x }{ 2 } ) } }{ (\frac { 3x }{ 2 } )^{ 2 } } } \times \dfrac { 9 }{ 4 } $$
$$\displaystyle =\frac{9}{2}$$
$$\Rightarrow \displaystyle \lim _{ x\rightarrow 0 }{ f\left( \frac { 1-cos3x }{ x^{ 2 } } \right) } = f\left( \dfrac 9 2\right)=\dfrac { 2}{ 9 } $$
.
The function $$f :( R-{0})$$ $$\rightarrow $$ R given by $$\displaystyle f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}$$ can be made continuous at $$x = 0$$ by defining $$f(0)$$ as
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$$2$$
0%
$$-1$$
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$$0$$
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$$1$$
Explanation
Given$$\displaystyle f\left( x \right) =\frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } $$
$$ \displaystyle \Rightarrow f\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \left\{ \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } \right\} } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 2x }-1-2x }{ x\left( { e }^{ 2x }-1 \right) } } \ ....... \quad \left[ \frac { 0 }{ 0 } form \right] $$
$$\therefore $$
usingL'Hospital rule
$$f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) } } $$
$$\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } } } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right) } =1$$
If the function $$f(x)= \left\{\begin{matrix}\dfrac {\sqrt {2+\cos x}-1}{(\pi-x)^2} & x\neq \pi \\ k & x=\pi \end{matrix}\right.$$ is continuous at $$x=\pi$$, then $$k$$ equals :
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$$0$$
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$$\dfrac {1}{2}$$
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$$2$$
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$$\dfrac {1}{4}$$
Explanation
$$\displaystyle\lim _{ x\rightarrow \pi }{ \frac { \sqrt { 2+\cos { x } } -1 }{ (\pi -x)^{ 2 } } } =\lim _{ x\rightarrow \pi }{ \frac { 1+\cos { x } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } } $$ [Rationalize numerator and denominator ]
$$\displaystyle =\lim _{ x\rightarrow \pi }{ \frac { 1-\cos { (\pi -x } ) }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } } $$
$$=\displaystyle \lim _{ x\rightarrow \pi }{ \frac { 2\sin ^{ 2 }{ \frac { (\pi -x) }{ 2 } } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } } =\frac { 1 }{ 4 } $$
So,$$ k =\dfrac{1}{4}$$ .
If $$f:R\rightarrow R$$ is a function defined by $$ f(x)=[x]\displaystyle \cos\left(\frac{2x-1}{2}\pi\right)$$, where $$[{x}]$$ denotes the greatest integer function, then $${f}$$ is:
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continuous for every real $$x$$
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discontinuous only at $$x = 0$$
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discontinuous only at non-zero integral values of $$x$$
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continuous only at $$x = 0$$
Explanation
For $$n\in I$$
$$\displaystyle\lim _{ x\rightarrow n+ }{ f\left( x \right) } =\lim _{ x\rightarrow n+ }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =n\cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0$$
And $$\displaystyle\lim _{ x\rightarrow n- }{ f\left( x \right) } =\lim _{ x\rightarrow n- }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =\left( n-1 \right) \cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0$$
Thus $$f$$ is continuous for $$x=n\in I$$ ........ (1)
Since the function $$g\left( x \right)=\left[ x \right] $$ and $$\displaystyle h\left( x \right) =\cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } $$ are continuous on $$x\in R-I$$, ......... (2)
From, (1) and (2), we get
$$f$$ is continuous everywhere
$$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$$
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0%
1
0%
0
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does not exist
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$$\infty$$
Explanation
$$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at} $$$$x\rightarrow 0$$
$$but\ have\ finite\ value = 0$$
$$0\times\ finite\ value\ =0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } } $$ is equal to
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$$3$$
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$$1$$
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$$0$$
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$$2$$
Explanation
$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } } $$
$$=\displaystyle \lim_{x\to 0}\dfrac{xe^x}{x}-\lim_{x\to 0}\dfrac{\sin x}{x}$$, split up
$$=\displaystyle \lim_{x\to 0}e^x-\lim_{x\to 0}\dfrac{\sin x}{x}$$
$$=e^0-1=1-1=0$$
Use limit properties to evaluate $$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x $$
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$$12$$
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$$14$$
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$$16$$
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$$18$$
Explanation
$$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x =\lim_{x\to 4}\left(3x\tan\dfrac{\pi}{x}\right)$$, simplify
$$=3(4)\tan\dfrac{\pi}{4}=12(1)=12$$, [substitute the limit ]
The from of the limit is not indeterminate, so we can substitute the limit value
For every integer $$n$$, let $$a_{n}$$ and $$b_{n}$$ be real numbers. Let function $$f: IR \rightarrow IR$$ be given by
$$f(x)=\left\{\begin{array}{l}
a_{n}+\sin\pi x,\ for\ x\in[2n,\ 2n+1]\\
b_{n}+\cos\pi x,\ for\ x\ \in(2n-1,2n)
\end{array}\right.$$$$, for\ all\ integers\ n.$$
lf $$f$$ is continuous, then which of the following hold(s) for all $$n$$?
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$$a_{n-1}-b_{n-1}=0$$
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$$a_{n}-b_{n}=1$$
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$$a_{n}-b_{n+1}=1$$
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$$a_{n-1}-b_{n}=-1$$
Explanation
Check the continuity at $$x=2n$$
$$L.H.L =\underset {x\rightarrow 2n^-}{\lim} [f(x)]$$
$$=\underset {h\rightarrow 0}{\lim}[f(2n-h)]$$
$$=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi (2n-h)]$$
$$=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi h]$$
$$=b_n+1$$
$$R.H.L=\underset {x\rightarrow 2n^+}{\lim}[f(x)]=\underset {h\rightarrow 0}{\lim}[f(2n+h)]$$
$$=\underset {h\rightarrow 0}{\lim}[a_n+\sin \pi(2n+h)]=a_n$$
$$f(2n)=a_n+\sin 2\pi n=a_n$$
For continuity, $$\underset {x\rightarrow 2n^-}{\lim}f(x)=\underset {x\rightarrow 2n^+}{\lim} f(x)=f(2n)$$
So, $$a_n=b_n+1\Rightarrow a_n-b_n=1$$
Now, check for continuity at $$x=2n+1$$
$$L.H.L.=\underset {h\rightarrow 0}{\lim}(a_n+\sin \pi (2n+1-h))=a_n$$
$$R.H.L.=\underset {h\rightarrow 0}{\lim}(b_{n+1}+\cos (\pi (2n+1-h)))=b_{n+1}-1$$
$$f(2n+1)=a_n$$.
For continuity, $$a_n=b_{n+1}-1\Rightarrow a_{n-1}=b_n-1$$
Both, option B and option D are correct.
If $$\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =A$$ and $$\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =B$$, then which one of the following is correct?
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$$A=1$$ and $$B=0$$
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$$A=0$$ and $$B=1$$
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$$A=0$$ and $$B=0$$
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$$A=1$$ and $$B=1$$
Explanation
As given
$$A=\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =\lim _{ x\rightarrow \infty }{ \cfrac { \sin { \left( \cfrac { 1 }{ x } \right) } }{ \left( \cfrac { 1 }{ x } \right) } } $$
Let $$t=\cfrac { 1 }{ x } ;x\rightarrow \alpha ,t\rightarrow 0$$
$$\Rightarrow A=\lim _{ t\rightarrow \infty }{ \cfrac { \sin { t } }{ t } } =1\left[ \because \lim _{ t\rightarrow 0 }{ \cfrac { \sin { x } }{ x } =1 } \right] $$
and $$B=\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } \Rightarrow B=0\quad $$
Therefore, $$A=1$$ and $$B=0$$ is correct
The function $$f\left( x \right)=\left[ x \right] ,$$ at $${ x }=5$$ is:
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left continuous
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right continuous
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continuous
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cannot be determined
Explanation
We know that the greatest integer function is discontinuous at integer values. For a value of $$x$$ just smaller than 5, the greatest integer smaller than it is 4. So,
$$\lim _{ x\rightarrow { 5 }^{ - } }{ [x] } = 4$$
$$f\left( x=5 \right)=\left[ 5 \right] =5$$
For a value of $$x$$ just bigger than 5, the greatest integer smaller than it is 5. So,
$$\lim _{ x\rightarrow { 5 }^{ + } }{ [x] } = 5$$
Since $$\lim _{ x\rightarrow { 5 }^{ - } }{ f\left( x \right) } \neq f\left( 5 \right)$$, the function is not left continuous.
Since $$\lim _{ x\rightarrow { 5 }^{ + } }{ f\left( x \right) } =f\left( 5 \right) $$, the function is right continuous.
Correct answer: Option B
$$f(x)=\left\{\begin{matrix} 2x-1& if &x>2 \\ k & if &x=2 \\ x^{2}-1& if & x<2\end{matrix}\right.$$is continuous at $$x= 2$$ then $$k =$$
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$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Since, the function is continuous at x = 2,
$$\displaystyle\lim_{ x \rightarrow {2}^{-}} f(x) = f(2) = \displaystyle\lim _{x \rightarrow {2}^{+} }f(x) $$
$$\displaystyle\lim _{x \rightarrow {2}^{-}} 2x - 1 = f(2) = \displaystyle\lim_{ x \rightarrow {2}^{+}} {x}^{2} - 1 $$
k = 3
Hence, k = 3
Evaluate $$\displaystyle \lim_{x \rightarrow -2} \displaystyle \frac{x^2 - 1}{2x + 4}$$.
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0
0%
1
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2
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$$\infty$$
Explanation
Consider, $$\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}$$
Substitute $$x=-2+h$$
$$\implies h\rightarrow 0 \text{ as } x\rightarrow -2$$
$$\implies \displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}$$
$$ \displaystyle =\lim _{ h\rightarrow 0 } \frac { (h-2)^{ 2 }-1 }{ 2(h-2)+4 } $$
$$ \displaystyle =\lim _{ h\rightarrow 0 } \frac { h^{ 2 }-4h+3 }{ 2h } $$
$$=\infty$$
Hence,
$$\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}=\infty$$
If $$f(x)=\displaystyle \frac{\sin x}{x},x\neq 0$$ is to be continuous at $${x}=0$$
then $$\mathrm{f}({0})=$$
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$$0$$
0%
$$1$$
0%
$$-1$$
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$$2$$
Explanation
From the property of limits,
$$\underset{ x \rightarrow 0}{Lim} \dfrac{sin x}{x} = 1 $$
Thus, for the function to be continuous at $$x = 0$$
$$f(0) = 1$$
Say true or false.
Every continuous function is always differentiable.
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True
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False
Explanation
If a function $$f(x)$$ is differentiable at a point $$a$$, then it is continuous at the point $$a$$. But the converse is not true.
$$f(x)=|x|$$ is contineous but not differentiable at $$x=0.$$
Therefore, the given statement is false.
$$\displaystyle \lim_{x\rightarrow \infty} \sin x$$ equals
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$$1$$
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$$0$$
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$$\infty$$
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does not exist
Explanation
Let $$L = \displaystyle \lim_{x\rightarrow \infty} \sin x$$
Assume $$\displaystyle y = \frac{1}{x}$$ so as $$x\to \infty, y \to 0$$
$$\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}$$
We know $$\sin x$$ lie between -1 to 1
so let $$p =\sin x$$ as $$x\to \infty$$
Thus left hand limit $$=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p$$
and right hand limit $$=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p$$
Clearly L.H.L $$\neq $$ R.H.L $$\Rightarrow$$ Required limit does not exist.
The function $$\displaystyle \mathrm{f}({x})=\frac{|x-3|}{x-3}$$ at $${x}=3$$, is
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Left continuous
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Right continuous
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continuous
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discontinuous
Explanation
we just concentrate on the function $$ \dfrac{|x|}{x} $$
LHL = lim $$ x \rightarrow {0}^{-} \dfrac{|x|}{x} = \dfrac{-x}{x} = -1 $$
RHL = lim $$ x \rightarrow {0}^{+} \dfrac{|x|}{x} = \dfrac{x}{x} = 1 $$
Let , $$y = x - 3$$
Now, if $$ x \rightarrow 3, then\ , y \rightarrow 0 $$
Applying the above derived result for y,
Since, the LHL is not equal to the RHL,
Hence the function is discontinuous at y = 0 or x = 3
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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