Limits and Derivatives Test

Result

Limits and Derivatives Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

the value of $$\underset { x\rightarrow 0 }{ lim } \frac { sin\alpha X-sin\beta X }{ { e }^{ \alpha X }-{ e }^{ \beta X } } $$ equals

A
0

B
1

C
-1

D
$$\alpha -\beta $$

Solution
Question 2
1 / -0

$$\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - 3\sin \left( {a + 2h} \right) + 3\sin \left( {a + h} \right) - \sin a}}{{{h^3}}}$$ is equal to

A
$$\cos a$$

B
$$ - \cos a$$

C
$$\sin a$$

D
$$\sin a\cos a$$

Solution
Question 3
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 4
1 / -0

$$L\underset { x\rightarrow \infty }{ im } \left( \sin { \sqrt { x+1 } -\sin { \sqrt { x } } } \right) =$$

A
2

B
-2

C
0

D
1
Question 5
1 / -0

$$ \lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{x \sin 2 x}= $$

A
1$$ / 2 $$

B
3$$ / 2 $$

C
3$$ / 4 $$

D
1$$ / 4 $$

Solution
Question 6
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 7
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 8
1 / -0

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

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

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

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

D
$$1$$

Solution
Question 9
1 / -0

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

A
2

B
-2

C
0

D
1

Solution
Question 10
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