Limits and Continuity Test 18

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Limits and Continuity Test 18

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Question 1
1 / -0

If $$\displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { n.{ 3 }^{ n } }{ n{ \left( x-2 \right) }^{ n }+n.{ 3 }^{ n+1 }-{ 3 }^{ n } } } =\cfrac { 1 }{ 3 } $$, then the range of $$x$$ is (When $$n\in N$$)

A
$$[2,\ 5)$$

B
$$\left( 1,\ 5 \right) $$

C
$$\left( -1,\ 5 \right) $$

D
$$\left( -\infty ,\ \infty \right) $$

Solution

$$\displaystyle\lim_{n\rightarrow \infty}\dfrac{n.3^n}{n(x-2)^n+n3^{n+1}-3^n}=\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{\left(\dfrac{x-2}{3}\right)^n+3-\dfrac{1}{n}}=\dfrac{1}{3}$$
We know $$\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{n}=0$$
then we want $$\dfrac{|x-2|}{3} < 1$$ $$\rightarrow \because$$ if $$a < 1, a^n=0$$ for $$n\rightarrow \infty$$
$$x < 5$$
$$x > -1$$
$$\therefore (-1, 5)$$.
Question 2
1 / -0

$$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}}$$ is equal to

A
$$1$$

B
$$3$$

C
$$\sqrt {3}$$

D
$$-\sqrt {3}$$

E
$$0$$

Solution

The value of $$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}} $$ is
$$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x} - \sqrt {3}}{\sqrt {x - 3}\sqrt {x} + 3} \times \dfrac {\sqrt {x} + \sqrt {3}}{\sqrt {x} + \sqrt {3}}$$
$$= \displaystyle \lim_{x\rightarrow 3} \dfrac {x - 3}{\sqrt {x - 3} \sqrt {x + 3}}\times \dfrac {1}{\sqrt {x} + \sqrt {3}}$$
$$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x - 3}}{\sqrt {x + 3}} \times \dfrac {1}{\sqrt {x} + \sqrt {3}} = 0$$
Question 3
1 / -0

What is $$\displaystyle \lim_{x \rightarrow 0 } x^2 \sin \left(\frac{1}{x}\right)$$ equal to ?

A
0

B
1

C
1/2

D
Limit does not exist.

Solution

$$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } } $$
$$={ 0 }^{ 2 }\times \sin { \left( \cfrac { 1 }{ 0 } \right) } \quad \left[ -1\le \sin { \left( \cfrac { 1 }{ x } \right) } \le 1 \right] $$
$$=0$$
Question 4
1 / -0

If $$f\left( x \right) =\begin{vmatrix} \sin { x } & \cos { x } & \tan { x } \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}$$, then $$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) }{ { x }^{ 2 } } } $$ is

A
$$-1$$

B
$$3$$

C
$$1$$

D
Zero

Solution

$$f(x)=\begin{vmatrix} sinx & cosx & tanx \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}\\ \Longrightarrow f(x)=sinx({ x }^{ 2 }-x)-cosx({ x }^{ 3 }-2{ x }^{ 2 })+tanx({ x }^{ 3 }-2{ x }^{ 3 })\\ \Longrightarrow f(x)={ x }^{ 2 }sinx-xsinx-{ x }^{ 3 }cosx+2{ x }^{ 2 }cosx-{ x }^{ 3 }tanx\\ \Longrightarrow \dfrac { f(x) }{ { x }^{ 2 } } =sinx-xcosx+2cosx-xtanx-\dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =\underset { x\longrightarrow 0 }{ lim } sinx-\underset { x\longrightarrow 0 }{ lim } xcosx+\underset { x\longrightarrow 0 }{ lim } 2cosx-\underset { x\longrightarrow 0 }{ lim } xtanx-\underset { x\longrightarrow 0 }{ lim } \dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =0-0+2-0-1\\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =1\\ \\ $$
Question 5
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } } $$ is

A
$$3$$

B
$$-3$$

C
$$6$$

D
$$0$$

Solution

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

So option $$D$$ is correct
Question 6
1 / -0

Which one of the following statements is correct?

A
$$\displaystyle \lim_{x \rightarrow 0} (fog) (x)$$ exists.

B
$$\displaystyle \lim_{x \rightarrow 0} (gof) (x)$$ exists.

C
$$\displaystyle \lim_{x \rightarrow 0-} (fog) (x) = \displaystyle \lim_{x \rightarrow 0-} (gof) (x)$$

D
$$\displaystyle \lim_{x \rightarrow 0+} (fog) (x) =\displaystyle \lim_{x \rightarrow 0-} (gof) (x)$$

Solution

$$(fog) (x) = [\sin x]$$
$$\displaystyle \lim_{x\to 0^+} [\sin x]=0$$ and $$\displaystyle \lim_{x\to 0^-} [\sin x]=-1$$. Hence, $$\displaystyle \lim_{x\to 0} fog(x)$$ does not exist.
Now, $$gof(x)=\sin [x]$$

$$\displaystyle \lim_{x\to 0^+} \sin [x]=0$$ and $$\displaystyle \lim_{x\to 0^-} \sin [x]=\sin (-1)$$. Hence, $$\displaystyle \lim_{x\to 0} gof(x)$$ does not exist.

But, $$\displaystyle \lim_{x\to 0^+} fog(x)=\displaystyle \lim_{x\to 0^-} gof(x)$$
Question 7
1 / -0

$$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$$ is equal to :

A
$$\pi+2$$

B
$$\pi-2$$

C
e

D
$$e^{-1}$$

Solution

$$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$$
$$=\pi - 2\left(\dfrac{\pi}{2}\right)^{\cos \frac{\pi}{2}}$$
$$=\pi-2\left(\dfrac{\pi}{2}\right)^{0}$$
$$=\pi-2$$
Question 8
1 / -0

$$\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}}$$ is equal to :

A
$$0$$

B
$$\displaystyle\frac{1}{(\sqrt{3}-2)}$$

C
$$1$$

D
$$\infty$$

Solution

$$ \lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \dfrac { \sin { (x-\cfrac { \pi }{ 6 } } ) }{ \sqrt { 3-2\cos { x } } } } $$
at $$\cfrac { \pi }{ 6 }$$ we get the numerator as $$\sin { (x-\cfrac { \pi }{ 6 } ) } = 0 $$
and the denominator as $$ \sqrt { 3-2\cos { x } } =\sqrt { 3-\sqrt { 3 } } $$
So,it is not an indeterminant form as we have $$\dfrac{0}{\sqrt{3 - \sqrt{3}} }$$
Hence the answer will be zero.
Question 9
1 / -0

If $$\underset{x\to 0}{\lim}\dfrac{x^a\sin^b x}{\sin(x^c)}, a, b, c, \in R \sim \{0\}$$ exists and has non-zero value, then

A
$$a,b,c$$ are in A.P

B
$$a,b,c$$ are in G.P

C
$$a,b,c$$ are in H.P

D
none of these

Solution

Given, limit is $$\lim_{x\rightarrow 0}\dfrac {x^a\sin^bx}{\sin (x^c)}$$
$$\underset{x\to 0}{\lim} x^a.\dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}.\dfrac{x^b}{x^c}$$
$$=\underset{x\to 0}{\lim} x^{a+b-c} \dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}$$
The above limit non-zero valu only when
$$a+b-c = 0$$
Question 10
1 / -0

Determine the value of k for which the following function is continuous at $$x=3$$.
$$f(x)=\dfrac{x^2-9}{x-3}, x \neq 3$$

$$f(x)=k, x=3$$

A
2

B
4

C
6

D
8

Solution

Since f(x) is continuous at $$x=3$$.

Therefore,

$$\displaystyle\lim_{x\rightarrow 3} f(x) =f(3)$$

$$\displaystyle\lim_{x\rightarrow 3} f(x)=k$$

$$\displaystyle\lim_{x\rightarrow 3} \dfrac{x^2-9}{x-3}=k$$

$$\displaystyle\lim_{x\rightarrow 3} \dfrac{(x+3)(x-3)}{x-3}=k$$

$$\displaystyle\lim_{x\rightarrow 3} (x+3)=k$$

$$k=6$$

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 18

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Limits and Continuity Test 18

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SHARING IS CARING

Question 1

$$[2,\ 5)$$

$$\left( 1,\ 5 \right) $$

$$\left( -1,\ 5 \right) $$

$$\left( -\infty ,\ \infty \right) $$

Solution

Question 2

$$1$$

$$3$$

$$\sqrt {3}$$

$$-\sqrt {3}$$

$$0$$

Solution

Question 3

0

1

1/2

Limit does not exist.

Solution

Question 4

$$-1$$

$$3$$

$$1$$

Zero

Solution

Question 5

$$3$$

$$-3$$

$$6$$

$$0$$

Solution

Question 6

$$\displaystyle \lim_{x \rightarrow 0} (fog) (x)$$ exists.

$$\displaystyle \lim_{x \rightarrow 0} (gof) (x)$$ exists.

$$\displaystyle \lim_{x \rightarrow 0-} (fog) (x) = \displaystyle \lim_{x \rightarrow 0-} (gof) (x)$$

$$\displaystyle \lim_{x \rightarrow 0+} (fog) (x) =\displaystyle \lim_{x \rightarrow 0-} (gof) (x)$$

Solution

Question 7

$$\pi+2$$

$$\pi-2$$

e

$$e^{-1}$$

Solution

Question 8

$$0$$

$$\displaystyle\frac{1}{(\sqrt{3}-2)}$$

$$1$$

$$\infty$$

Solution

Question 9

$$a,b,c$$ are in A.P

$$a,b,c$$ are in G.P

$$a,b,c$$ are in H.P

none of these

Solution

Question 10

2

4

6

8

Solution

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