Limits and Continuity Test 23

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Limits and Continuity Test 23

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Question 1
1 / -0

$$\lim\limits_{x\rightarrow0}\dfrac{x\tan 2x-2x\tan x}{\left(1-\cos 2x\right)^{2}}=$$

A
$$2$$

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

C
$$-2$$

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

Solution

$$\underset{x\rightarrow 0}{\lim}\dfrac{x\tan2x-2x\tan x}{(1-\cos 2x)^2}$$
$$\underset{x\rightarrow 0}{\lim}\dfrac{x[\tan2x-2\tan x]}{(1-\cos 2x)^2}$$
$$\underset{x\rightarrow 0}{\lim}x\dfrac { \left[ \frac { 2\tan { x } }{ 1-{ \tan { x } }^{ 2 } } -2\tan { x } \right] }{ { \left( 2{ \sin { x } }^{ 2 } \right) }^{ 2 } } $$
$$\underset{x\rightarrow 0}{\lim}{ \dfrac { 2x\tan { x\left[ \dfrac { 1 }{ 1-{ \tan { x } }^{ 2 } } -1 \right] } }{ 4{ \sin { x } }^{ 4 } } } $$
$$\underset{x\rightarrow 0}{\lim}\dfrac { \dfrac { x\tan { x } }{ \left( 1-{ \tan { x } }^{ 2 } \right) } \left[ 1-1+{ \tan { x } }^{ 2 } \right] }{ 2{ \sin { x } }^{ 4 } } $$
$$\underset{x\rightarrow 0}{\lim} {\dfrac { x{ \tan^3 { x } }^{ 3 } }{ ( 1-{ \tan^2 { x } )}.2{ \sin^4 { x } } } } $$
Divide num and den by $$x^4$$
$$\underset{x\rightarrow 0}{\lim}\dfrac{\dfrac{\sin^3 x}{\cos^3 x}.\dfrac{x}{x^4}}{2\dfrac{\sin^4x}{x^4}(1-\tan^2 x)}$$
$$\underset{x\rightarrow 0}{\lim}{ \dfrac { { \left( \dfrac { \sin { x } }{ x } \right) }^{ 3 }\dfrac { 1 }{ { \cos ^{ 3 }{ x } } } }{ 2{ \left( \dfrac { \sin { x } }{ x } \right) }^{ 4 }\left( 1-{ \tan^2 { x } } \right) } } $$
Apply limit
$${ \dfrac { { \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 3 }.\dfrac { 1 }{ { \cos ^{ 3 }{1 } } } }{ 2{ \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 4 }\left( 1-{ \tan^2 { 1 } }\right) } }=\dfrac{1.1}{2.1(1-0)}$$
$$=\dfrac{1.1}{2.1(1-0)}$$
$$=\dfrac{1}{2}$$
$$\therefore \underset{x\rightarrow 0}{\lim}\dfrac{x \tan 2x- 2x\tan x}{(1-\cos 2x)^2}=\dfrac{1}{2}$$
Question 2
1 / -0

The value of $$\displaystyle lim_{x\to 0} \dfrac{cos (sin x) - cos x}{x^4} $$ is equal to

A
$$1/5$$

B
$$1/6$$

C
$$1/4$$

D
$$1/2$$

Solution

Given,

$$\lim _{x\to \:0}\left(\dfrac{\cos \left(\sin \left(x\right)\right)-\cos \left(x\right)}{x^4}\right)$$

apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)+\sin \left(x\right)}{4x^3}\right)$$

again apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{-\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)+\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos \left(x\right)}{12x^2}\right)$$

once again apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{\frac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{2}}{24x}\right)$$

upon simplification, we get,

$$=\lim _{x\to \:0}\left(\dfrac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{48x}\right)$$

again applying L-Hospital's rule,

$$=\lim _{x\to \:0}\left(\dfrac{3\left(2\cos \left(2x\right)\cos \left(\sin \left(x\right)\right)-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)\sin \left(2x\right)\right)+2\left(-3\cos ^2\left(x\right)\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^4\left(x\right)\cos \left(\sin \left(x\right)\right)\right)+2\left(-\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)\right)-2\cos \left(x\right)}{48}\right)$$

substituting $$x=0$$

$$\dfrac{3\left(2\cos \left(2\cdot \:0\right)\cos \left(\sin \left(0\right)\right)-\sin \left(\sin \left(0\right)\right)\cos \left(0\right)\sin \left(2\cdot \:0\right)\right)+2\left(-3\cos ^2\left(0\right)\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^4\left(0\right)\cos \left(\sin \left(0\right)\right)\right)+2\left(-\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^2\left(0\right)\cos \left(\sin \left(0\right)\right)\right)-2\cos \left(0\right)}{48}$$

we get,

$$=\dfrac{1}{6}$$
Question 3
1 / -0

$$\lim_{x\to \infty} \dfrac{\sqrt{x^2 + sin^2x}}{x+cosx}$$ equals

A
$$1$$

B
$$0$$

C
$$\infty$$

D
does not exist

Solution

Given,

$$\lim _{x\to \infty \:}\left(\dfrac{\sqrt{x^2+\sin ^2\left(x\right)}}{x+\cos \left(x\right)}\right)$$

dividing by highest denominator power

$$=\lim _{x\to \infty \:}\left(\dfrac{\sqrt{1+\frac{\sin ^2\left(x\right)}{x^2}}}{1+\frac{\cos \left(x\right)}{x}}\right)$$

$$=\dfrac{\lim _{x\to \infty \:}\left(\sqrt{1+\frac{\sin ^2\left(x\right)}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{\cos \left(x\right)}{x}\right)}$$

$$=\dfrac{\sqrt{1+0}}{1+0}$$

$$=1$$
Question 4
1 / -0

$$\underset{x\rightarrow 0}{lim} \dfrac{\sqrt{a^2 -ax+x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} -\sqrt{a-x}}$$ is equal to (a > 0)

A
$$\sqrt{a}$$

B
$$-\sqrt{a}$$

C
$$a \sqrt{a}$$

D
$$-a\sqrt{a}$$

Solution

$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\sqrt{a}}{2x}(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\left(\dfrac{(a^2-ax+x^2)-(a^2+ax+x^2)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\right)$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\times \dfrac{(-2ax)}{2\sqrt{a}}=-\sqrt{a}$$
$$\Rightarrow (B)$$.
Question 5
1 / -0

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

A
$$ \dfrac{1}{24} $$

B
$$ \dfrac{1}{16} $$

C
$$ \dfrac{1}{8} $$

D
$$ \dfrac{1}{4} $$
Question 6
1 / -0

$$\lim _ { x \rightarrow 1 } \{ 1 - x + [ x + 1 ] + [ 1 - x ] \} , \text { where } [ x ]$$ denotes greatest integer function is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution
Question 7
1 / -0

$$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}=$$__________.

A
$$\cos\sqrt{x}$$

B
$$\dfrac{1}{2\sin \sqrt{x}}$$

C
$$\dfrac{\cos \sqrt{x}}{2\sqrt{x}}$$

D
$$\sin\sqrt{x}$$

Solution

$$Lt_{x\rightarrow 0}\cfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{(x+h)-x}$$
Let $$f(x)=\sin\sqrt{x}$$
$$Lt_{h\rightarrow 0}\cfrac{\sin{x+h}-\sin\sqrt{x}}{(x+h)-x}=Lt_{x\rightarrow 0}f\cfrac{(x+h)-f(x)}{h}=f^1(x)\\ \Rightarrow \cos\sqrt x\times\cfrac{1}{2\sqrt x}\\ \Rightarrow \cfrac{\cos\sqrt x}{2\sqrt x}$$
Question 8
1 / -0

$$ \underset { x\rightarrow 0 }{ lim } \cfrac { 1+cos\left( \pi x \right) }{ \left( 1-x \right)^ 2 } $$ is equal to :

A
$$ \cfrac { \pi} {2} $$

B
$$ - \cfrac { \pi ^2} {2} $$

C
$$\cfrac { \pi ^2} {2} $$

D
$$ \cfrac { \pi} {3} $$

Solution
Question 9
1 / -0

Evaluate: $$\underset{x\rightarrow 0}{lim} \dfrac{e^{1/x} - 1}{e^{1/x }+ 1}$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
does not exist

Solution

As it is $$ \dfrac{\infty }{\infty }$$ form, applying L'hospital

$$ L = \lim_{x\rightarrow 0}\dfrac{{e^{1/x}(-1/x^{2})}}{{e^{1/x}(-1/x^{2})}} = 1 \Rightarrow (B) $$
Question 10
1 / -0

If $$\underset { x\rightarrow { 0 } }{ lim } \dfrac { \left( ax+b \right) -\sqrt { 4+\sin x } }{ \tan\quad x } =\dfrac { 27 }{ 4 } ~where ~a,b\in R$$ then the value of

A
$$a = 2~ and~ b = 7$$

B
$$a = -2~and~ b = 7 $$

C
$$a = 7~ and~ b = 2 $$

D
$$a=7~ and ~b = -2$$

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Limits and Continuity Test 23

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Limits and Continuity Test 23

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SHARING IS CARING

Question 1

$$2$$

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

$$-2$$

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

Solution

Question 2

$$1/5$$

$$1/6$$

$$1/4$$

$$1/2$$

Solution

Question 3

$$1$$

$$0$$

$$\infty$$

does not exist

Solution

Question 4

$$\sqrt{a}$$

$$-\sqrt{a}$$

$$a \sqrt{a}$$

$$-a\sqrt{a}$$

Solution

Question 5

$$ \dfrac{1}{24} $$

$$ \dfrac{1}{16} $$

$$ \dfrac{1}{8} $$

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

Question 6

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 7

$$\cos\sqrt{x}$$

$$\dfrac{1}{2\sin \sqrt{x}}$$

$$\dfrac{\cos \sqrt{x}}{2\sqrt{x}}$$

$$\sin\sqrt{x}$$

Solution

Question 8

$$ \cfrac { \pi} {2} $$

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

$$\cfrac { \pi ^2} {2} $$

$$ \cfrac { \pi} {3} $$

Solution

Question 9

$$0$$

$$1$$

$$-1$$

does not exist

Solution

Question 10

$$a = 2~ and~ b = 7$$

$$a = -2~and~ b = 7 $$

$$a = 7~ and~ b = 2 $$

$$a=7~ and ~b = -2$$

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