Limits and Continuity Test 25

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Limits and Continuity Test 25

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } $$ where $$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$$, is

A
$$2$$

B
$$\dfrac {1}{6}$$

C
$$\dfrac {2}{3}$$

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

Solution

$$\mathop {f\left( x \right)}\limits_{\lim \,\,x \to 0} = \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}$$
$$ = \frac{{ - \sin \left( {\sin x} \right)\cos x - \sin x}}{{4{x^3}}}$$
$$ = \frac{{ - \left( {\sin x} \right){{\cos }^2}x + \sin \left( {\sin x} \right)\sin x - \cos x}}{{12{x^2}}}$$
$$ = \frac{{\sin \left( {\sin x} \right){{\cos }^3}x + \cos \left( {\sin x} \right)\cos 2x + \cos \left( {\sin x} \right)\sin x + \sin \left( {\sin x} \right)\cos x + \sin x}}{{24x}}$$
$$ = \frac{{1 + 1 + Q + 1 + 1}}{{24}} = \frac{1}{6}$$
Question 2
1 / -0

$$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } \int _{ 0 }^{ x }{ \left( \sqrt { { t }^{ 2 }+5t } -t \right) dt } }$$

A
$$0$$

B
$$1$$

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

D
$$5$$

Solution
Question 3
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } } \left( a>b \right) $$ is

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

B
$$\dfrac {1}{a\sqrt {a-b}}$$

C
$$\dfrac {1}{2a\sqrt {a-b}}$$

D
$$\dfrac {1}{4a\sqrt {a-b}}$$

Solution

We have,
$$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } }$$

This is the $$\dfrac{0}{0}$$ form.
Apply L-hospital rule,
$$\lim_{x\to a}\dfrac{\dfrac{1}{2\sqrt {x-b}}-0}{2x-0}$$
$$\lim_{x\to a}\dfrac{1}{4x\sqrt {x-b}}$$
$$=\dfrac{1}{4a\sqrt {a-b}}$$

Hence, this is the answer.
Question 4
1 / -0

$$\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }$$ equals :

A
$$- 1 / 6$$

B
$$1 / 6$$

C
$$1 / 3$$

D
$$- 1 / 3$$

Solution

$$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 27+x \right) }^{ 1/3 }-3 }{ 9-{ \left( 27+x \right) }^{ 2/3 } } $$
$$=\underset { x\rightarrow 0 }{ lt } \dfrac { \dfrac { 1 }{ 3 } { \left( 27+x \right) }^{ -2/3 } }{ \dfrac { -2 }{ 3 } { \left( 27+x \right) }^{ -1/3 } } $$
$$=\underset { x\rightarrow 0 }{ lt } \dfrac { -1 }{ 2 } \dfrac { { \left( 27+x \right) }^{ 1/3 } }{ { \left( 27+x \right) }^{ 2/3 } } $$
$$=\dfrac { -1 }{ 2 } \left( \dfrac { 3 }{ 9 } \right) ,\quad \dfrac { -1 }{ 6 } $$ [A]
Question 5
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x }{ 5 } } \left[ \frac { x }{ 2 } \right] $$ (where $$[.]$$ denotes the greatest integer function) is

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

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

C
$$0$$

D
$$\infty$$

Solution

Given,

$$\lim _{x\to 0}\left(\dfrac{x}{5}\left[\dfrac{x}{2}\right]\right)$$

$$\lim _{x\to \:0}\left(\dfrac{x}{5}\cdot \dfrac{x}{2}\right)$$

put the value $$x=0$$

$$=\dfrac{0}{5}\cdot \dfrac{0}{2}$$

$$=0$$
Question 6
1 / -0

$$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } } \right] }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right) }{ { x }^{ 3 }-{ x }^{ -1 } } \right] =$$

A
$$\frac { 1 }{ 3 } $$

B
3

C
$$\frac { 1 }{ 2 } $$

D
$$\frac { 3 }{ 2 } $$

Solution

$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4}{x^2-x^{-1}}-\dfrac{1-3x+x^2}{1-x^3}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$$
$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4x}{x^3-1}+\dfrac{1-3x+x^2}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$$
$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{x^2+x+1}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)x}{x^4-1}\right]$$
$$\displaystyle\lim_{x\rightarrow 0}\left[\dfrac{(x-1)(x^2+x+1)}{x^2+x+1}+3x\right]$$ as $$(x^3-1)2(x-1)(x^2+x+1)$$
$$\displaystyle\lim_{x\rightarrow 1}[(x-1)+3x]=(1-1)+3(1)=3$$
Answer$$=3$$.
Question 7
1 / -0

$$\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x } $$ is :

A
$$-2$$

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

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

D
$$2$$

Solution

Let $$L=\displaystyle \lim_{x\rightarrow 0} \frac { (1 - cos 2x)^2}{2x \tan x - x \tan 2x}= \lim_{x\rightarrow 0} \frac {( 2 sin x)^2}{2x \tan x- \frac {x 2 \tan x}{1 - \tan^2 x}}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac { { 4 sin^4 x}}{\frac{1 - \tan^2 x}{2 x \tan x - 2x \tan^3 x - 2x \tan x}}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac {4 sin ^4 x ( 1 - \tan^2 x)}{- 2 x \tan^3 x}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac {2 sin^4 x}{-x \frac {sin^3 x}{cos x}}( 1 - \tan^2 x)$$

$$\displaystyle =- 2 \lim_{x\rightarrow 0} \left \{ \frac {sin x . cos^3 x}{x} (1 \tan^2 x) \right \}$$

$$ \displaystyle = -2 \left ( \lim_{x\rightarrow 0} \frac { sin x}{x} \right ) \lim_{x\rightarrow 0} \left \{ cos^3 x ( 1 - \tan^2 x) \right \}$$

$$L\displaystyle = -2(1)(1)$$

$$L\displaystyle = -2$$
Question 8
1 / -0

The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}$$ is

A
$$10/3$$

B
$$3/10$$

C
$$6/5$$

D
$$5/6$$

Solution

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right) \cdot \sin 5x}}{{{x^2}\sin 3x}}$$
$$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ 2{{\sin }^2}x \cdot \sin 5x}}{{{x^2}\sin 3x}}$$
$$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin x}}{x}} \right)}^2} \cdot \left( {\dfrac{{\sin 5x}}{{5x}}} \right) \cdot 5x}}{{\left( {\dfrac{{\sin 3x}}{{3x}}} \right) \cdot 3x}}$$
$$ = \dfrac{{10}}{3}$$.
Question 9
1 / -0

$$\underset { x\rightarrow 0 }{ \lim } \dfrac { { 3 }^{ 2x }-{ 2 }^{ 3x } }{ x } $$ is equal to

A
$$\log\dfrac { 3 }{ 2 } $$

B
$$1$$

C
$$\log\cfrac { 9 }{ 8 } $$

D
$$0$$

Solution

We have,

$$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}-{{2}^{3x}}}{x}$$

Applying L’ Hospital rule and we get,

$$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}\log 3\times2-{{2}^{3x}}\log 2\times(3)}{1}$$

Taking limit and we get,

$$ ={{3}^{2\times 0}}\log 3^2-{{2}^{3\times 0}}\log 2^3 $$

$$ =\log 9-\log 8 $$

$$ =\log \dfrac{9}{8} $$

Hence, this is the answer.
Question 10
1 / -0

Let $$f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2$$ and $$lim_{x\rightarrow \infty} f(x)=1$$ then $$f(-2)=$$

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$0$$

Solution

$$\underset{x\rightarrow 0}{lim} f(x)=2$$
$$\Rightarrow \underset{x\rightarrow 0}{lim} \dfrac{ax+b}{x+1}=2$$
$$\Rightarrow \dfrac{b}{1}=2$$
$$\therefore b=2$$

Its is also given that
$$\underset{x\rightarrow \infty}{lim} f(x)=1$$
$$\Rightarrow \underset{x\rightarrow \infty}{lim} \dfrac{ax+b}{x+1}=1$$

$$\Rightarrow \underset{x\rightarrow \infty}{lim}\dfrac{a+\dfrac{b}{x}}{1+\dfrac{1}{x}}=1$$

$$\Rightarrow \dfrac{a+0}{1+0}=1$$

$$\therefore a=1$$
Substituting the values of a and b in $$f(x)=\dfrac{ax+b}{x+1}$$, we get,

$$f(x)=\dfrac{x+2}{x+1}$$

$$\therefore f(-2)=\dfrac{-2+2}{-2+1}=0$$

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Limits and Continuity Test 25

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Limits and Continuity Test 25

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SHARING IS CARING

Question 1

$$2$$

$$\dfrac {1}{6}$$

$$\dfrac {2}{3}$$

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

Solution

Question 2

$$0$$

$$1$$

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

$$5$$

Solution

Question 3

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

$$\dfrac {1}{a\sqrt {a-b}}$$

$$\dfrac {1}{2a\sqrt {a-b}}$$

$$\dfrac {1}{4a\sqrt {a-b}}$$

Solution

Question 4

$$- 1 / 6$$

$$1 / 6$$

$$1 / 3$$

$$- 1 / 3$$

Solution

Question 5

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

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

$$0$$

$$\infty$$

Solution

Question 6

$$\frac { 1 }{ 3 } $$

3

$$\frac { 1 }{ 2 } $$

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

Solution

Question 7

$$-2$$

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

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

$$2$$

Solution

Question 8

$$10/3$$

$$3/10$$

$$6/5$$

$$5/6$$

Solution

Question 9

$$\log\dfrac { 3 }{ 2 } $$

$$1$$

$$\log\cfrac { 9 }{ 8 } $$

$$0$$

Solution

Question 10

$$1$$

$$2$$

$$-1$$

$$0$$

Solution

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