Limits and Continuity Test 34

Result

Limits and Continuity Test 34

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$,
then the value of $$\displaystyle\lim_{x\rightarrow a}{f(x)g(x)}$$ is?

A
Does not exist

B
Exists and is $$\displaystyle\frac{3}{4}$$

C
Exists and is $$\displaystyle-\frac{3}{4}$$

D
Exists and is $$\displaystyle\frac{4}{3}$$

Solution

Given, $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$,

Lets, assume $$F(x) = (f(x) + g(x))$$ and $$ G(x) = (f(x) - g(x))$$
$$\because$$ Limits of both $$F(x)$$ and $$G(x)$$ exists as $$x\rightarrow a$$, we can say that $$\displaystyle\lim_{x\rightarrow a}{(F(x)+G(x))}$$ and $$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))}$$ also exists.

$$\displaystyle\therefore\lim_{x\rightarrow a}{(F(x)+G(x))} = 3$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{f(x)} = 3$$
$$\Rightarrow \lim_{x\rightarrow a}{f(x)} = \dfrac{3}{2}$$
Similarly,
$$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))} = 1$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{g(x)} = 1$$
$$\Rightarrow \lim_{x\rightarrow a}{g(x)} = \dfrac{1}{2}$$
Hence,
$$\displaystyle\lim_{x\rightarrow a}{(f(x)\cdot g(x))} = \lim_{x\rightarrow a}{f(x)}\cdot\lim_{x\rightarrow a}{g(x)} = \dfrac{3}{2}\times \dfrac{1}{2} = \dfrac{3}{4}$$

Hence, option B.
Question 2
1 / -0

$$x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
$$f(x)$$ $$4$$ $$3$$ $$7$$ $$1$$ $$3$$
The function f is continuous on the closed interval $$[1, 5]$$ and values of the function are shown in the table above. If the values in the table are used to calculate a trapezoidal sum, the approximate value of $$\int_{1}^{5}f(x)dx$$ is

A
$$14$$

B
$$14.5$$

C
$$15$$

D
$$29$$

Solution
Question 3
1 / -0

The value of $$\displaystyle \underset { n\rightarrow \infty }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n } \right) $$ is

A
$$\displaystyle \log { 2 } $$

B
$$\displaystyle \log { 6 } $$

C
$$\displaystyle 1$$

D
$$\displaystyle \log { 3 } $$
Question 4
1 / -0

A point where function $$f(x)$$ is not continuous where $$f(x)=\left[ \sin { \left[ x \right] } \right] $$ in $$\left( 0,2\pi \right) $$; is ($$\left[ \ast \right] $$ denotes greatest integer $$\le x$$)

A
$$(3,0)$$

B
$$(2,0)$$

C
$$(1,0)$$

D
$$(4,-1)$$
Question 5
1 / -0

The value of $$\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x } \right) }^{ 1/x }+{ \left( 1+x \right) }^{ \sin { x } } \right) } $$ whre $$x> 0$$ is

A
$$0$$

B
$$-1$$

C
$$1$$

D
2

Solution

Given $$\lim _{ x\rightarrow 0 }{ \left( { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+{ \left( 1+x \right) }^{ sinx } \right) } $$

$$=\lim _{ x\rightarrow 0 }{ { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } } $$

$$={ \left( sin0 \right) }^{ { \frac { 1 }{ 0 } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } $$

$$={ \left( 0 \right) }^{ { \infty } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } $$

$$=0+{ e }^{ \lim _{ x\rightarrow 0 }{ \left[ log{ \left( 1+x \right) }^{ sinx } \right] } }$$
$$\left( \because { e }^{ loga }=a \right) $$

$$={ e }^{ \lim _{ x\rightarrow 0 }{ \left[ sinx.log{ \left( 1+x \right) } \right] } }$$

$$={ e }^{ \lim _{ x\rightarrow 0 }{ sinx } \times \lim _{ x\rightarrow 0 }{ log\left( 1+x \right) } }$$

$$={ e }^{ sin\left( 0 \right) \times log\left( 1+0 \right) }$$

$$={ e }^{ 0\times log\left( 1 \right) }$$

$$={ e }^{ 0\times 0 }={ e }^{ 0 }$$

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

$$ \underset { n \rightarrow \infty }{ Lt } \sum _{ r=2n+1\quad }^{ 3n } \dfrac {n}{r^2 - n^2} $$ is equal to :

A
$$ l n \sqrt{\frac {2}{3}} $$

B
$$ l n \sqrt{\frac {3}{2}} $$

C
$$ l n \frac {2}{3} $$

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

Solution
Question 7
1 / -0

The value of $$\displaystyle \lim_{x \rightarrow 1^{-}}\dfrac {1 - \sqrt {x}}{(\cos^{-1}x)^{2}}$$

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

B
$$1/2$$

C
$$2$$

D
None of these

Solution

$${ lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) \\ { \Rightarrow lim }_{ x\rightarrow 1 }1-\sqrt { x } =0\quad ,\quad { \Rightarrow lim }_{ x\rightarrow 1 }(cos^{ -1 }x)^{ 2 }=0\\ { lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) =\frac { 0 }{ 0 } \\ Now\quad using\quad l'hospital's\quad rule\\ { lim }_{ x\rightarrow 1 }\left( \dfrac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) ={ lim }_{ x\rightarrow 1 }\left( \dfrac { \frac { d(1-\sqrt { x } ) }{ dx } }{ \frac { d(cos^{ -1 }x)^{ 2 } }{ dx } } \right) ={ lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } \\ Now\quad { lim }_{ x\rightarrow 1 }1-\dfrac { 1 }{ 2\sqrt { x } } =1-1/2=1/2\\ And\quad { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } =\frac { 0 }{ 0 } \\ Again\quad using\quad l'hospital's\quad rule\\ \\ { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2d(cos^{ -1 }x)/dx }{ d(\sqrt { 1-{ x }^{ 2 } } )/dx } ={ lim }_{ x\rightarrow 1 }\dfrac { -2(\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } )\quad }{ \frac { -2x }{ 2\sqrt { 1-{ x }^{ 2 } } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2 }{ x } =-2\\ Therefore,\quad { lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } =\dfrac { 1/2 }{ -2 } =-1/4\\ $$
Question 8
1 / -0

$$\displaystyle \lim_{I\rightarrow \left (\dfrac {\pi}{2}\right )} = \int_{0}^{t}\tan \theta \sqrt {\cos \theta} ln (\cos \theta) d\theta$$ is equal to

A
$$-4$$

B
$$4$$

C
$$-2$$

D
Does not exists
Question 9
1 / -0

If $$f '$$ (0) = 0 and f(x) is a differentiable and increasing function,then lim $$ x \rightarrow 0$$ $$\frac {x.f ' (x^2)}{f ' (x)}$$

A
is always equal to zero

B
may not exist as left hand limit may not exist

C
may not exist as left hand limit may not exist

D
right hand limit is always zero
Question 10
1 / -0

$$\quad \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \sum _{ r=1 }^{ 2n }{ \cfrac { r }{ \sqrt { { n }^{ 2 }+{ r }^{ 2 } } } } } $$ equal to:

A
$$1+\sqrt { 5 } $$

B
$$-1+\sqrt { 5 } $$

C
$$1+\sqrt { 2 } $$

D
$$1+\sqrt { 2 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$x$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
$$f(x)$$	$$4$$	$$3$$	$$7$$	$$1$$	$$3$$

Limits and Continuity Test 34

Result

Limits and Continuity Test 34

Score

-

Rank

-

SHARING IS CARING

Question 1

Does not exist

Exists and is $$\displaystyle\frac{3}{4}$$

Exists and is $$\displaystyle-\frac{3}{4}$$

Exists and is $$\displaystyle\frac{4}{3}$$

Solution

Question 2

$$14$$

$$14.5$$

$$15$$

$$29$$

Solution

Question 3

$$\displaystyle \log { 2 } $$

$$\displaystyle \log { 6 } $$

$$\displaystyle 1$$

$$\displaystyle \log { 3 } $$

Question 4

$$(3,0)$$

$$(2,0)$$

$$(1,0)$$

$$(4,-1)$$

Question 5

$$0$$

$$-1$$

$$1$$

2

Solution

Question 6

$$ l n \sqrt{\frac {2}{3}} $$

$$ l n \sqrt{\frac {3}{2}} $$

$$ l n \frac {2}{3} $$

$$ l n \frac {3}{2} $$

Solution

Question 7

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

$$1/2$$

$$2$$

None of these

Solution

Question 8

$$-4$$

$$4$$

$$-2$$

Does not exists

Question 9

is always equal to zero

may not exist as left hand limit may not exist

may not exist as left hand limit may not exist

right hand limit is always zero

Question 10

$$1+\sqrt { 5 } $$

$$-1+\sqrt { 5 } $$

$$1+\sqrt { 2 } $$

$$1+\sqrt { 2 } $$

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.