Limits and Continuity Test 41

Result

Limits and Continuity Test 41

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Question 1
1 / -0

The value of $$\displaystyle\lim_{x\to 0} |x|^{sinx}$$ equals

A
$$0$$

B
$$-1$$

C
$$1$$

D
does not exist

Solution
Question 2
1 / -0

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

A
$$\frac { 3 } { 2 }$$

B
$$\frac { 5 } { 2 }$$

C
$$\frac { 7 } { 4 }$$

D
$$\frac { 9 } { 2 }$$
Question 3
1 / -0

Arrange the following limits in the ascending order :
(1) $$\lim _ { x \rightarrow \infty } \left( \dfrac { 1 + x } { 2 + x } \right) ^ { x + 2 }$$

(2) $$\lim _ { x \rightarrow 0 } ( 1 + 2 x ) ^ { 3 / x }$$

(3) $$\lim _ { \theta \rightarrow 0 } \dfrac { \sin \theta } { 2 \theta }$$

(4) $$\lim _ { x \rightarrow 0 } \dfrac { \log _ { e } ( 1 + x ) } { x }$$

A
$$1,2,3,4$$

B
$$1,3,4,2$$

C
$$1,4,3,2$$

D
$$3,4,1,2$$

Solution

(i) $$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { 1+x }{ 2+x } \right) }^{ x+2 }$$
$$=$$ $${ \left( \dfrac { \dfrac { 1 }{ x } +1 }{ \dfrac { 2 }{ x } +1 } \right) }^{ x+2 }$$
$$=\underset { x\rightarrow \infty }{ lim } { e }^{ ln{ \left( \dfrac { 1+x }{ 2+x } \right) }^{ 2+x } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \left( 2+x \right) ln\left( \dfrac { 1+x }{ 2+x } \right) }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { ln\left( 1+x \right) -ln\left( 2+x \right) }{ \left( 1/2+x \right) } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \dfrac { 1 }{ 1+x } -\dfrac { 1 }{ 2+x } }{ \dfrac { -1 }{ { \left( 2+x \right) }^{ 2 } } } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { { \left( 2+x \right) }^{ 2 }\left[ \left( 2+x \right) -\left( 1+x \right) \right] }{ \left( 2+x \right) -\left( 1+x \right) \left( 2+x \right) \left( 1+x \right) } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \left( 2+x \right) }{ 1+x } }$$
$$={ e }^{ 1 }$$
(ii) $$\underset { x\rightarrow 0 }{ lim } { \left( 1+2x \right) }^{ 3/x }$$
$$={ e }^{ \underset { x\rightarrow 0 }{ lt } \left( 2x \right) 3/x }$$
$$={ e }^{ 6 }$$
(iii) $$\underset { \theta \rightarrow 0 }{ lt } \dfrac { \sin\theta }{ 2\theta } =\underset { \theta \rightarrow 0 }{ lt } \dfrac { \cos\theta }{ 2 } =\dfrac { 1 }{ 2 } $$
(iv) $$\underset { x\rightarrow 0 }{ lt } \dfrac { { log }_{ e }\left( 1+x \right) }{ x } .\underset { x\rightarrow 0 }{ lt } \dfrac { 1 }{ 1+x } =1$$
$$\therefore$$ $$3<4<1<2$$.
Question 4
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 5
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 6
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 7
1 / -0

$$\mathop{\lim}\limits_{x \to 0} \left(\dfrac{3+x}{3-x}\right)^{\dfrac{x+1}{x}}$$ is equal to

A
$$e^{2/3}$$

B
$$e^{1/3}$$

C
$$e^3$$

D
$$e^2$$

Solution
Question 8
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 9
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 10
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