Limits and Continuity Test 22

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Limits and Continuity Test 22

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Weekly Quiz Competition

Question 1
1 / -0

$$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} = $$

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

B
$$\dfrac{1}{2}$$

C
$$2$$

D
$$-2$$

Solution
Question 2
1 / -0

$$\mathop {\lim }\limits_{x \to 0} \,\dfrac{1}{x}\,{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to

A
$$1$$

B
$$0$$

C
$$2$$

D
$$\dfrac{1}{2}$$

Solution

Solution -
$$ \displaystyle \lim_{x\rightarrow 0} \frac{1}{x}sin^{-1}(\frac{2x}{1+x^{2}})$$
we know that
$$ \displaystyle sin^{-1}x+sin^{-1}y = sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})$$
$$ \displaystyle sin^{-1}(\frac{2x}{1+x^{2}})= sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})+sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})$$
$$ \displaystyle = 2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi $$
$$\displaystyle \lim_{x\rightarrow 0} \frac{2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi }{x}$$
from triangle.
$$ \displaystyle \lim_{x\rightarrow 0} \frac{2tan^{-1}(\frac{1}{x})-\pi }{x}$$
Apply L Hospital rule-
$$ \displaystyle \lim_{x\rightarrow 0} 2(\frac{1}{1+(\frac{1}{x})^{2}})$$
$$ \displaystyle \lim_{x\rightarrow 0} \frac{2x^{2}}{1+x^{2}}$$
$$ = 0.$$
B is correct.
Question 3
1 / -0

solve the limit
$$\mathop {\lim }\limits_{x \to 3} \dfrac{2}{{x - 3}}$$

A
$$2$$

B
$$3$$

C
$$4$$

D
Does not exist

Solution

$$\displaystyle \lim_{x\rightarrow 3} \dfrac{2}{x-3}$$
put $$h = x - 3 $$
$$\displaystyle \lim_{h\rightarrow 0} \dfrac{2}{h}$$
this is $$\dfrac{2}{0}$$ that is not defined.
so limit does not exist.
Question 4
1 / -0

Solve

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}$$

A
$$-\dfrac{5}{3}$$

B
$$\dfrac{5}{3}$$

C
$$\dfrac{7}{3}$$

D
None of these

Solution

Let$$L=\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}$$

$$= \displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}\times \dfrac {3x}{\tan x}\times \dfrac {5}{3}$$

we know that $$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}=1$$ and $$\displaystyle \lim_{x\rightarrow 0}\dfrac {3x}{\tan 3x}=1$$

$$L=1\times 1\times \dfrac {5}{3}$$

$$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}= \dfrac {5}{3}$$
Question 5
1 / -0

$$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to

A
$$1$$

B
$$0$$

C
$$2$$

D
$$1/2$$

Solution

$$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$

can be written as
$$\mathop {\lim }\limits_{x \to 0} \dfrac{\sin^{-1}(\dfrac{2x}{1+x^2})}{x}$$

it is of the form $$\dfrac00$$

So, we will apply L Hospital rule until we will get determinate form

Differentiating numerator and denominator we will get,
$$=\lim_{x\to 0}\dfrac{\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}}{1}=\lim_{x\to0}\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}=2$$
Question 6
1 / -0

$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}} = $$

A
$$\dfrac{1}{{10}}$$

B
$$\dfrac{{ - 1}}{{10}}$$

C
$$\dfrac{2}{5}$$

D
$$ - \dfrac{2}{5}$$

Solution

Now,
$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(2{x^2} +3 x-2x - 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(x-1)(2x + 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(\sqrt x-\sqrt 1)(\sqrt x+\sqrt 1)(2x + 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)}}{{(\sqrt x+\sqrt 1)(2x + 3)}} $$

$$=\dfrac{-1}{2\times5}$$

$$=-\dfrac{1}{10}$$.
Question 7
1 / -0

If $${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$$, where $$ r= 1, 2, 3, ....n$$, then $$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right)$$ is equal to

A
$$\cos \alpha + i\sin \alpha $$

B
$$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$$

C
$${e^{i\alpha /2}}$$

D
$$\sqrt {{e^{i\alpha }}} $$

Solution

Solution -
$$ z_{r} = \dfrac{cosr\alpha }{n^{2}}+i\dfrac{sinr\alpha }{n^{2}} $$

Comparing with $$ cos \theta +isin \theta$$

$$\theta =\dfrac {r\alpha}{n^2}$$

$$\therefore z_r= e^{\dfrac{ir\alpha }{n^{2}}}$$

$$ \lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = e^{\dfrac{i\alpha }{n^{2}}}e^{\dfrac{2i\alpha }{n^{2}}} e^{\dfrac{3i\alpha }{n^{2}}}...e^{\dfrac{ni\alpha }{n^{2}}}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{1}{n^{2}}+\dfrac{2}{n^{2}}+...)}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha \dfrac{1}{n^2}(1+2+3)}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{n^{2}+n}{2n^{^{2}}})} \quad \dots S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$$

at $$ n\rightarrow \infty $$

$$ = e^{i\alpha }\times \dfrac{1}{2}$$

$$\lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = \sqrt{e^{i\alpha }}$$
Question 8
1 / -0

Solve:
$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=$$

A
$$\dfrac{4}{3}(\sqrt{2}+1)$$

B
$$\dfrac{4}{3}(\sqrt{2}-1)$$

C
$$\dfrac{3}{4}(\sqrt{2}-1)$$

D
$$\dfrac{3}{4}(\sqrt{2}-2)$$

Solution

$$\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}}$$

$$=\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}\times\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}$$

$$=\displaystyle\int_{0}^{1}{\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}}$$

$$=\displaystyle\int_{0}^{1}{\left(\sqrt{x+1}+\sqrt{x}\right)dx}$$

$$=\left[\dfrac{{\left(x+1\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+\dfrac{{\left(x\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$$

$$=\left[2\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+2\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$$

$$=2\left[\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$$

$$=\dfrac{2}{3}\left[\left({2}^{\frac{3}{2}}-1\right)-\left(1-0\right)\right]$$

$$=\dfrac{2}{3}\left[2\sqrt{2}-2\right]$$

$$=\dfrac{4}{3}\left(\sqrt{2}-1\right)$$
Question 9
1 / -0

If $$f(x)$$ is the integral of $$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$$. Find $$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right) } $$, where $$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$$

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

B
$$1$$

C
$$\dfrac{1}{3}$$

D
$$2$$

Solution

$$f(x)=\dfrac{2\sin x-\sin 2x}{x^3}$$ $$x\neq 0$$

$$f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}$$

$$=\dfrac{2\sin x}{x}\left(\dfrac{1-\cos x}{x^2}\right)$$

$$\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)$$

$$\Rightarrow \displaystyle 4\cdot 1 \lim_{x\rightarrow 0}{\left(\dfrac{\sin^2(x/2)}{4\times (x/2)^2}\right)}$$

$$=1 $$
Question 10
1 / -0

Find the value of limit $$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 } }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 } } = } $$.

A
$$0$$

B
$$3$$

C
$$-3$$

D
$$1$$

Solution

$$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +2\sin { x } -\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -2\sin { x } -\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin { x } \left( \sin { x } +1 \right) -1\left( \sin { x } +1 \right) }{ 2\sin { x } \left( \sin { x } -1 \right) -1\left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \left( 2\sin { x } -1 \right) \left( \sin { x } +1 \right) }{ \left( 2\sin { x } -1 \right) \left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \sin { x } +1 }{ \sin { x } -1 } } \\ =\frac { \frac { 1 }{ 2 } +1 }{ \frac { 1 }{ 2 } -1 } \\ =-3$$

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Limits and Continuity Test 22

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Limits and Continuity Test 22

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SHARING IS CARING

Question 1

$$\dfrac{-1}{2}$$

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

$$2$$

$$-2$$

Solution

Question 2

$$1$$

$$0$$

$$2$$

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

Solution

Question 3

$$2$$

$$3$$

$$4$$

Does not exist

Solution

Question 4

$$-\dfrac{5}{3}$$

$$\dfrac{5}{3}$$

$$\dfrac{7}{3}$$

None of these

Solution

Question 5

$$1$$

$$0$$

$$2$$

$$1/2$$

Solution

Question 6

$$\dfrac{1}{{10}}$$

$$\dfrac{{ - 1}}{{10}}$$

$$\dfrac{2}{5}$$

$$ - \dfrac{2}{5}$$

Solution

Question 7

$$\cos \alpha + i\sin \alpha $$

$$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$$

$${e^{i\alpha /2}}$$

$$\sqrt {{e^{i\alpha }}} $$

Solution

Question 8

$$\dfrac{4}{3}(\sqrt{2}+1)$$

$$\dfrac{4}{3}(\sqrt{2}-1)$$

$$\dfrac{3}{4}(\sqrt{2}-1)$$

$$\dfrac{3}{4}(\sqrt{2}-2)$$

Solution

Question 9

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

$$1$$

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

$$2$$

Solution

Question 10

$$0$$

$$3$$

$$-3$$

$$1$$

Solution

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