Limits and Derivatives Test

Result

Limits and Derivatives Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$lim_{x \to 0} (\dfrac{1}{x^2} - cotx)$$ equals

A
$$1$$

B
$$0$$

C
$$\infty$$

D
Does not exist

Solution
Question 2
1 / -0

$$ \displaystyle \lim _{ x-\infty }{ sgn\left( \cot{\dfrac { { \pi x }^{ 2019 } }{ { x }^{ 2019 }+7 }} \right) }$$

A
Equals $$-1$$

B
Equals $$1$$

C
equals $$0$$

D
Does not exit

Solution
Question 3
1 / -0

$$\displaystyle\lim_{x \to \pi/2} (sec x +tan x)$$ is equal to

A
$$1$$

B
$$-1$$

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

D
$$0$$

Solution
Question 4
1 / -0

$$\underset { x\rightarrow 0 }{ lim } \dfrac { x\tan { 2x } -2\tan { 2x } }{ { \left( 1-cos2x \right) } }$$ equals:

A
$$\dfrac{1}{4}$$

B
$$1$$

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

D
$$-\dfrac{1}{2}$$
Question 5
1 / -0

$$\underset{x \rightarrow 0}{Lt} \dfrac{sin x - x + x^3 / 6}{x^5}$$ =

A
$$\dfrac{1}{120}$$

B
$$\dfrac{1}{110}$$

C
$$\dfrac{1}{100}$$

D
none of these

Solution
Question 6
1 / -0

$$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$$ is equal to

A
$$\cos^{4}\alpha.\log(\cos\alpha)-\sin^{4}\alpha.\log(\sin\alpha)$$

B
$$\sin^{4}\alpha.\log(\cos\alpha)-\cos^{4}\alpha.\log(\sin\alpha)$$

C
$$\sin^{4}\alpha.\log(\cos\alpha)+\cos^{4}\alpha.\log(\sin\alpha)$$

D
$$None\ of\ the \ above$$

Solution
Question 7
1 / -0

$$\displaystyle\lim_{x\to \pi/2} \dfrac{sinx-(sinx)^{sin x}}{1-sin x + In sin x}$$ is equal to

A
$$4$$

B
$$2$$

C
$$1$$

D
none of these

Solution
Question 8
1 / -0

$$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$$ is equal to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 9
1 / -0

If $$\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 1,$$ then

A
$$a = \frac{5}{2}$$

B
$$b = \frac{{ - 5}}{2}$$

C
$$a + b = 4$$

D
$$a + b = -4$$

Solution
Question 10
1 / -0

Evaluate : $$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right) }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x } }{ { e }^{ x\ell nx } } \right) } $$ is equal to

A
$$\dfrac{1}{e}\ell n3$$

B
$$e\ \ell n\ 3$$

C
$$3$$

D
$$\dfrac{1}{3}$$

Solution