Limits and Continuity Test 53

Result

Limits and Continuity Test 53

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Weekly Quiz Competition

Question 1
1 / -0

$$\lim _ { x \rightarrow 0 } \frac { \ln ( \sin 3 x ) } { \ln ( \sin x ) }$$ is equal to

A
0

B
1

C
2

D
none of these

Solution
Question 2
1 / -0

$$\displaystyle {Lt}_{x\rightarrow 0}\dfrac{cos5x cos3x}{x(sin5x sin3x)}$$ =

A
$$-4$$

B
$$4$$

C
$$-\dfrac{1}{4}$$

D
$$\dfrac{1}{4}$$

Solution
Question 3
1 / -0

$$\underset { x\rightarrow \infty }{ Lt } { 5 }^{ x }sin\left( \cfrac { a }{ { 5 }^{ x } } \right) =$$

A
0

B
5

C
log5

D
None of these.

Solution
Question 4
1 / -0

The value of $$\theta ,\quad is$$
$$\underset { o\rightarrow o }{ lim } \quad \frac { { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\quad \left( 1-{ cos }^{ 2 }\quad .....\left( { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\theta \right\} \right) \right) \right\} }{ sin\left( \frac { \pi (\sqrt { \theta +4 } -2 }{ \theta } \right) } $$

A
$$\frac { \sqrt { 2 } }{ 4 } $$

B
$$\sqrt { 2 } $$

C
1

D
2

Solution
Question 5
1 / -0

$$\underset { x\rightarrow \infty }{ Lim } (sin\sqrt { x+1 } -sin\sqrt { x } )=$$

A
2

B
-2

C
0

D
1

Solution
Question 6
1 / -0

$$\lim _{ x\rightarrow \infty }{ \frac { \sin { x } \sin { (\frac { \pi }{ 3 } +x) } \sin { (\frac { \pi }{ 3 } -x) } }{ x } } =$$

A
$$\frac { 3 }{ 4 } $$

B
$$\frac { 1 }{ 4 } $$

C
$$\frac { 4 }{ 3 } $$

D
$$0$$

Solution
Question 7
1 / -0

$$\lim _{ x\rightarrow 0 }{ \frac { { 27 }^{ x }-{ 9 }^{ x }-{ 3 }^{ x }+1 }{ \sqrt { 2 } -\sqrt { 1+\cos { x } } } = } $$

A
$$0$$

B
$$8\sqrt { 2 } { (\log { 3 } ) }^{ 2 }$$

C
$$8{ (\log { 3 } ) }^{ 2 }$$

D
$$1$$

Solution
Question 8
1 / -0

$$\lim _{ x\rightarrow -\infty }{ \cfrac { x^{ 4 }\sin { \cfrac { 1 }{ x } } +x^{ 2 } }{ 1+\left| x \right| ^{ 3 } } = } $$

A
1

B
-1

C
2

D
-3

Solution
Question 9
1 / -0

$$\lim _{ x\rightarrow }{ 0 } \{ (sinx-x)/{ x }^{ 3 })\} $$ equals:

A
$$1/3$$

B
$$-1/3$$

C
$$1/6$$

D
$$-1/6$$

Solution
Question 10
1 / -0

Solve
$$\underset { x\rightarrow 0 }{ Lim } \frac { 8 }{ { x }^{ 8 } } \left( 1-cos\frac { { x }^{ 2 } }{ 2 } -cos\frac { { x }^{ 2 } }{ 4 } +cos\cfrac { { x }^{ 2 } }{ 2 } .cos\cfrac { { x }^{ 2 } }{ 4 } \right) =$$

A
$$\cfrac { 1 }{ 16 } $$

B
$$\cfrac { 1 }{ 15 } $$

C
$$\cfrac { 1 }{ 32 } $$

D
$$1$$

Solution