Limits and Continuity Test 49

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Limits and Continuity Test 49

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Weekly Quiz Competition

Question 1
1 / -0

Let [x] denote the greatest integer less than or equal to x. Then :
$\underset { x\rightarrow 0 }{ lim } \dfrac { tan\left( \pi { sin }^{ 2 }x \right) +\left( \left| x \right| -sin \right) \left( x\left[ x \right] \right) ^{ 2 } }{ { x }^{ 2 } } :$

A
does not exist

B
equal $\pi$

C
equal 0

D
equal $\pi$ + 1
Question 2
1 / -0

$\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } \frac { cotx-cosx }{ { (\pi -2x) }^{ 3 } }$ equals :

A
$\frac { 1 }{ 8 }$

B
$\frac { 1 }{ 4 }$

C
$\frac { 1 }{ 24 }$

D
$\frac { 1 }{ 16 }$

Solution
Question 3
1 / -0

$\displaystyle\lim _{ n\rightarrow \infty }{ { n }^{ 2 }\left( { x }^{ \dfrac { 1 }{ n } }-{ x }^{ \dfrac { 1 }{ n+1 } } \right) ,x>0 }$ is equal to

A
$0$

B
$e^{x}$

C
$log_ex$

D
$None\ of\ these$

Solution
Question 4
1 / -0

If $\displaystyle \lim _{ x\rightarrow 0 }[1+ax+bx^{2}]^{(2/x)}=e^{3}$ , then

A
$a=3,\ b=0$

B
$a=\dfrac{3}{2},\ b\neq1$

C
$a=\dfrac{3}{2},\ b=R$

D
$a=2,\ b=3$

Solution
Question 5
1 / -0

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

A
$1$

B
$-1$

C
$2$

D
$-2$

Solution
Question 6
1 / -0

$\displaystyle x\xrightarrow { lim } a\left(\sin\frac{x-a}{2}\tan\frac{\pi x}{2a} \right)$

A
$\displaystyle a/\pi$

B
$-\displaystyle a/\pi$

C
$\displaystyle \pi/a$

D
$-\displaystyle \pi/a$

Solution
Question 7
1 / -0

$\lim _{ x\rightarrow 0 }{ \frac { \sqrt [ 3 ]{ 1+\sin { x } } -\sqrt [ 3 ]{ 1-\sin { x } } }{ x } } =$

A
$0$

B
$1$

C
$\frac { 2 }{ 3 }$

D
$\frac { 3 }{ 2 }$

Solution
Question 8
1 / -0

$\underset { x\rightarrow 0 }{ lim } \frac { sin({ 6x }^{ 2 }) }{ Incos({ 2x }^{ 2 }-x) } =$

A
12

B
-12

C
6

D
-6

Solution
Question 9
1 / -0

$\underset { x\rightarrow -\infty }{ lim } \frac { ({ 3x }^{ 4 }+{ 2x }^{ 2 })sin(\frac { 1 }{ x } )+{ |x| }^{ 3 }+5 }{ { |x }|^{ 3 }+{ |x| }^{ 2 }+|x|+1 } =$

A
2

B
1

C
-2

D
-3
Question 10
1 / -0

The value of $\displaystyle \lim_{x \rightarrow 0} (\sin x)^{\dfrac{1}{x}}+(1+x)^{(\sin x)})=0$ , where $x > 0$ , is :

A
$0$

B
$-1$

C
$1$

D
$2$

Solution

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 49

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Limits and Continuity Test 49

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Question 1

does not exist

equal π\pi π

equal 0

equal π\pi π + 1

Question 2

18\frac { 1 }{ 8 } 81​

14\frac { 1 }{ 4 } 41​

124\frac { 1 }{ 24 } 241​

116\frac { 1 }{ 16 } 161​

Solution

Question 3

000

exe^{x}ex

logexlog_exloge​x

None of theseNone\ of\ theseNone of these

Solution

Question 4

a=3, b=0a=3,\ b=0a=3, b=0

a=32, b≠1a=\dfrac{3}{2},\ b\neq1a=23​, b=1

a=32, b=Ra=\dfrac{3}{2},\ b=Ra=23​, b=R

a=2, b=3a=2,\ b=3a=2, b=3

Solution

Question 5

111

−1-1−1

222

−2-2−2

Solution

Question 6

a/π\displaystyle a/\pia/π

−a/π-\displaystyle a/\pi−a/π

π/a\displaystyle \pi/aπ/a

−π/a-\displaystyle \pi/a−π/a

Solution

Question 7

000

111

23\frac { 2 }{ 3 } 32​

32\frac { 3 }{ 2 } 23​

Solution

Question 8

12

-12

6

-6

Solution

Question 9

2

1

-2

-3

Question 10

000

−1-1−1

111

222

Solution

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Question Analysis

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equal $\pi$

equal $\pi$ + 1

$\frac { 1 }{ 8 }$

$\frac { 1 }{ 4 }$

$\frac { 1 }{ 24 }$

$\frac { 1 }{ 16 }$

$0$

$e^{x}$

$log_ex$

$None\ of\ these$

$a=3,\ b=0$

$a=\dfrac{3}{2},\ b\neq1$

$a=\dfrac{3}{2},\ b=R$

$a=2,\ b=3$

$1$

$-1$

$2$

$-2$

$\displaystyle a/\pi$

$-\displaystyle a/\pi$

$\displaystyle \pi/a$

$-\displaystyle \pi/a$

$0$

$1$

$\frac { 2 }{ 3 }$

$\frac { 3 }{ 2 }$

$0$

$-1$

$1$

$2$