Limits and Derivatives Test

Result

Limits and Derivatives Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

If $$ \lim _{x \rightarrow 0}\left(\cos x+a^{3} \sin \left(b^{6} x\right)\right)^{\frac{1}{x}}=e^{512} $$ then value of $$ab^2$$ is equal to

A
$$-512$$

B
$$512$$

C
$$8$$

D
none of these

Solution
Question 2
1 / -0

$$if\left( x \right) =\left[ x-3 \right] +\left[ x-4 \right] \quad for\quad x\epsilon R\quad then\quad \underset { x\rightarrow 3 }{ lim } f\left( x \right) =$$

A
-2

B
-1

C
0

D
2

Solution
Question 3
1 / -0

$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin(\pi \cos^2x)}{x^2}$$ is equal to?

A
$$-\pi$$

B
$$\pi$$

C
$$\pi/2$$

D
$$1$$

Solution
Question 4
1 / -0

$$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { { \left( \cos { x } \right) }^{ 1/2 }-{ \left( \cos { x } \right) }^{ 1/3 } }{ \sin ^{ 2 }{ x } } } $$ is $$

A
$$1/6$$

B
$$-1/12$$

C
$$2/3$$

D
$$1/3$$

Solution
Question 5
1 / -0

$$\displaystyle\lim_{x\rightarrow 0}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$$ is equal to?

A
$$2$$

B
$$-2$$

C
$$1/2$$

D
$$-1/2$$

Solution
Question 6
1 / -0

If $$f(x)$$ is odd linear polynomial with $$f(1)=1$$, then $$\underset{x \to 0} {\lim} \dfrac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}$$ is

A
$$1$$

B
$$\ln 2$$

C
$$\dfrac{1}{2}\ln 2$$

D
$$\cos 2$$

Solution
Question 7
1 / -0

$$\lim _ { x \rightarrow \infty } \dfrac { \sin \left( \pi \cos ^ { -2 } x \right) } { x ^ { 2 } }$$ is equal to:

A
$$- \pi$$

B
$$ \pi$$

C
$$\dfrac { \pi } { 2 }$$

D
$$1$$

Solution
Question 8
1 / -0

$$\lim _ { x \rightarrow 0 } \dfrac { \sin 4 x } { \tan 7 x } =$$

A
$$1$$

B
$$\dfrac { 4 } { 7 }$$

C
$$\dfrac { 7 } { 4 }$$

D
$$0$$

Solution
Question 9
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{x(1+a\cos x)-b\sin x}{x^{3}}}=1$$ then

A
$$a=-5/2$$

B
$$a=-3/2,b=-1/2$$

C
$$a=-3/2,b=-5/2$$

D
$$a=-5/2,b=-3/2$$

Solution
Question 10
1 / -0

$$\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?$$

A
$$2_{\displaystyle /3}$$

B
$$0$$

C
$$1_{\displaystyle /3}$$

D
$$-2_{\displaystyle /3}$$

Solution