Limits and Derivatives Test

Result

Limits and Derivatives Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

$$\lim _ { x \rightarrow 0 } \dfrac { | \cos ( \sin ( 3 x ) ) | - 1 } { x ^ { 2 } }$$ equals

A
$$\dfrac { - 9 } { 2 }$$

B
$$\dfrac { - 3 } { 2 }$$

C
$$\dfrac { 3 } { 2 }$$

D
$$\dfrac { 9 } { 2 }$$

Solution
Question 2
1 / -0

Consider $$\lim _ { x \rightarrow 0 } \dfrac { a x + b e ^ { - x } + \sin x + 1 } { a x - b \sin x } = \ell ( \ell$$ is a finite number)

A
$$\ell = 0 , \text { if } a = - 2 , b = - 1$$

B
$$\ell = 5 , i f a = 1 , b = - 1$$

C
$$\ell = 1 , \text { if } a = b = - 1$$

D
None of these

Solution
Question 3
1 / -0

$$\lim _ { x \rightarrow 0 } \dfrac { \sin x ^ { 5 } } { \sin ^ { 4 } x } =$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$\infty$$

Solution
Question 4
1 / -0

$$\lim _ { x \rightarrow 0 } \left( \left[ \dfrac { - 5 \sin x } { x } \right] + \left[ \dfrac { 6 \sin x } { x } \right] \right)$$ (where $$[ .]$$ denotes greatest integer function) is equal to

A
$$0$$

B
$$-12$$

C
$$1$$

D
$$2$$

Solution
Question 5
1 / -0

$$\displaystyle\lim_{x\rightarrow \dfrac{\pi}{4}}\dfrac{\cos x-\sin x}{\left(\dfrac{\pi}{4}-x\right)(\cos x+\sin x)}=?$$

A
$$2$$

B
$$1$$

C
$$0$$

D
$$3$$

Solution
Question 6
1 / -0

$$\displaystyle\lim_{x\rightarrow 0}\dfrac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}=?$$

A
$$10/3$$

B
$$3/10$$

C
$$6/5$$

D
$$56$$

Solution
Question 7
1 / -0

$$\lim _ { x \rightarrow - 1 } \dfrac { \cos 2 - \cos 2 x } { x ^ { 2 } - | x | }$$ is equal to :

A
$$0$$

B
$$\cos 2$$

C
$$2 \sin 2$$

D
None

Solution
Question 8
1 / -0

$$\lim _{ x\rightarrow 0 }{ \cfrac { x.{ 10 }^{ x }-x }{ 1-cosx } = } $$

A
$$\log { 10 } $$

B
$$2\log { 10 } $$

C
$$3\log { 10 } $$

D
$$4\log { 10 } $$

Solution
Question 9
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 10
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