Limits and Continuity Test 30

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Limits and Continuity Test 30

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle\lim_{n\rightarrow \infty}\dfrac{(n 1)^{1/n}}{n}$$ equals?

A
$$e$$

B
$$e^{-1}$$

C
$$1$$

D
None of these

Solution
Question 2
1 / -0

If $$ f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}}, $$ then

A
$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

B
$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

C
$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

D
none of these

Solution

d. $$\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \dfrac{-\sin t}{(1-\cos t)^{1 / 3}}$$
$$=-\lim _{t \rightarrow 0} \dfrac{2 \sin \dfrac{t}{2} \cos \dfrac{t}{2}}{\left(2 \sin ^{2} \dfrac{t}{2}\right)^{1 / 3}}$$
$$ =-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \dfrac{t}{2}\left(\sin \dfrac{t}{2}\right)^{1 / 3}=0 $$
Question 3
1 / -0

The value of $$ \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x} $$ is

A
1/3

B
2/3

C
-1/4

D
3/2

Solution

We have $$ \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x} $$
$$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{(1+\cos x)\left(1-\cos x+\cos ^{2} x\right)}{(1-\cos x)(1+\cos x)}$$
$$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{1-\cos x+\cos ^{2} x}{1-\cos x}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}$$
Question 4
1 / -0

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n) $$ is equal to

A
1

B
0

C
n/m

D
None of these

Solution

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}}=\lim _{x \rightarrow 0}\left(\dfrac{\sin x^{n}}{x^{n}}\right)\left(\dfrac{x^{n}}{x^{m}}\right)\left(\dfrac{x}{\sin x}\right)^{m} $$
$$ =\displaystyle \lim _{x \rightarrow 0} x^{n-m}=0 \quad[\because m<n] $$
Question 5
1 / -0

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$ is equal to

A
0

B
1

C
2

D
4

Solution

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$
$$ =\displaystyle \lim _{x \rightarrow 1} \dfrac{1-\cos \left(\dfrac{3 \pi}{2}-\dfrac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)} $$
$$ =\lim _{x \rightarrow 1} \dfrac{2 \sin ^{2}\left(\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}\right)}{2 \sin ^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\left(\dfrac{\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}}{\dfrac{\pi}{2}-\dfrac{\pi x}{2}}\right)^{2} $$
$$ =\displaystyle \lim _{x \rightarrow 1} 9\left(\dfrac{\dfrac{1}{2}-\dfrac{x}{1+x^{2}}}{1-x}\right)^{2}=\lim _{x \rightarrow 1} 9\left(\dfrac{x-1}{2\left(1+x^{2}\right)}\right)^{2}=0 $$
Question 6
1 / -0

The value of $$ \displaystyle \lim _{x \rightarrow 2} \dfrac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}} $$ is

A
16

B
8

C
4

D
2

Solution

$$ \displaystyle \lim _{x \rightarrow 2} \dfrac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}$$
$$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}\right)^{2}-6 \times 2^{x}+2^{3}}{\sqrt{2^{x}}-2}\left[\text { Multiplying } N^{r} \text { and } D^{r} \text { by } 2^{x}\right]$$
$$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}-4\right)\left(2^{x}-2\right)(\sqrt{2^{x}}+2)}{(\sqrt{2^{x}}-2)(\sqrt{2^{x}}+2)} $$
$$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}-4\right)\left(2^{x}-2\right)(\sqrt{2^{x}}+2)}{\left(2^{x}-4\right)}$$
$$=\displaystyle \lim _{x \rightarrow 2}\left(2^{x}-2\right)(\sqrt{2^{x}}+2)=\left(2^{2}-2\right)(2+2)=8$$
Question 7
1 / -0

$$ The \ value \ of \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$ is

A
$$e^{-2 \pi} $$

B
$$ e^{1 / \pi} $$

C
$$ e^{2 /\pi} $$

D
$$ e^{-1 / \pi} $$

Solution

$$ \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \dfrac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \cot \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1} \dfrac{(1-x)}{\tan \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}} $$
$$=e^{\dfrac{2}{\pi}\displaystyle \lim_{x \to 1}{\dfrac{\dfrac{\pi}{2}(1-x)}{tan \left(\dfrac{\pi}{2}(1-x)\right)}}}$$
$$=e^{\dfrac{2}{\pi}}$$
Question 8
1 / -0

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x} \text { is equal to }$$

A
$$ \dfrac{1}{2 \pi} $$

B
$$ \dfrac{-1}{\pi} $$

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

D
None of these

Solution

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \dfrac{1+x}{2 \pi}=\dfrac{-1}{\pi}$$
Question 9
1 / -0

$$ \displaystyle \lim _{x \to \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$ is equal to

A
1

B
-1

C
0

D
$$None \ of \ these$$

Solution

$$ \displaystyle \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$

$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{2 x \sin x-\pi}{2 \cos x} \quad\quad\quad \left(\dfrac{0}{0}form\right)$$

$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{[2 \sin x+2 x \cos x]}{-2 \sin x} $$
(Applying L'Hopital's rule) $$ =-1 $$
Question 10
1 / -0

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$ is equal to

A
1

B
0

C
2

D
None of these

Solution

$$ \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$
$$ =\dfrac{x^{4}\left(1+\tan ^{2} x+\tan ^{4} x\right)}{\tan ^{4} x\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\dfrac{x^{4}}{\tan ^{4} x}, x \neq 0 $$
$$ \Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}}{\tan ^{4} x}=1 $$

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 30

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Limits and Continuity Test 30

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SHARING IS CARING

Question 1

$$e$$

$$e^{-1}$$

$$1$$

None of these

Solution

Question 2

$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

none of these

Solution

Question 3

1/3

2/3

-1/4

3/2

Solution

Question 4

1

0

n/m

None of these

Solution

Question 5

0

1

2

4

Solution

Question 6

16

8

4

2

Solution

Question 7

$$e^{-2 \pi} $$

$$ e^{1 / \pi} $$

$$ e^{2 /\pi} $$

$$ e^{-1 / \pi} $$

Solution

Question 8

$$ \dfrac{1}{2 \pi} $$

$$ \dfrac{-1}{\pi} $$

$$ \dfrac{-2}{\pi} $$

None of these

Solution

Question 9

1

-1

0

$$None \ of \ these$$

Solution

Question 10

1

0

2

None of these

Solution

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