Limits and Continuity Test 19

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Limits and Continuity Test 19

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Weekly Quiz Competition

Question 1
1 / -0

$$\lim _{ x\rightarrow 3 }{ \left( { x }^{ 3 }-4 \right) /\left( x+1 \right) } =$$

A
$$(4/23)$$

B
$$(2/23)$$

C
$$(1/8)$$

D
$$(23/4)$$
Question 2
1 / -0

Evaluate the limit:
$$\displaystyle \lim_{x \to 0} \left( \frac{x- \sin x}{x}\right) \sin \left(\frac{1}{x} \right)$$

A
$$1$$

B
$$-1$$

C
Does not exist

D
$$0$$

Solution

$$\lim _{ x\rightarrow 0 }{ (\cfrac { x-\sin { x } }{ x } ) } \sin { (\cfrac { 1 }{ x } ) } \\ \lim _{ x\rightarrow 0 }{ (\cfrac { x-\sin { x } }{ x } ) } \times \lim _{ x\rightarrow 0 }{ \sin { (\cfrac { 1 }{ x } ) } } \\ (1-\cos { x } )\times (Value\quad between\quad -1\quad and\quad 1)\\ 0\times Value\quad between\quad (-1\& 1)\\ 0$$
Question 3
1 / -0

$$\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x } \right) }^{ x }+{ \left( \cos { x } \right) }^{ \frac { 1 }{ \ln { x } } }+{ \left( x\sin { x } \right) }^{ x } \right) } $$ is equal to

A
$$2$$

B
$$2+e$$

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

D
$$3$$

Solution

$$\lim_{x\rightarrow 0^+}(x\cos x)^x$$

$$\lim_{x\rightarrow 0^+} x^x.\cos ^x x$$

$$\lim_{x\rightarrow 0^+} e^{x\ln x}.\cos ^x x$$

$$ 1\times 1$$

$$1$$

$$\therefore \lim_{x\rightarrow 0^+}(x\cos x)^x=1$$
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$$\lim_{x\rightarrow 0^+}(x\sin x)^x$$

$$\lim_{x\rightarrow 0^+} x^x.\sin ^x x$$

$$\lim_{x\rightarrow 0^+} e^{x\ln x}.\sin ^x x$$

$$ 1\times 1$$

$$1$$

$$\therefore \lim_{x\rightarrow 0^+}(x\sin x)^x=1$$

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$$\lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}$$

$$\lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$$

$$y= \lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$$

$$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\ln x}(-\ln (\sin x))$$

$$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\frac{1}{x}}\left ( -\frac{\cos x}{\sin x} \right )$$

$$\ln y =\lim_{x\rightarrow 0^+}(-\cos x)$$

$$\ln y = -1$$

$$y=e^{-1}$$

$$y=\dfrac{1}{e}$$

$$\therefore \lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}=\dfrac{1}{e}$$

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$$\lim_{x\rightarrow 0^+} ((x\cos x)^x+(\csc x)^{\frac{1}{\ln x}}+(x\sin x)^x)$$

$$=1+\dfrac{1}{e}+1$$

$$=2+\dfrac{1}{e}$$
Question 4
1 / -0

If the function $$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$$ for $$x \neq 0$$ continuous at $$x=0$$ then $$f(0)=$$

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

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

C
$$2$$

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

Solution

Here , function is continuous ,

So, left limit and right limit boh are equal and equals to the value of $$f(x)$$ at $$x=0$$

Therefore

$$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}-cosx}{x^2}$$

Solving limit by derivative approach,

=> $$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}.2x+sinx}{2x}$$

=> $$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{2e^{x^2}+e^{x^2}.x^2+cosx}{2x}$$

Putting $$x=0$$, we get

=> $$f(0)=\dfrac{2+0+1}{2}=\dfrac{3}{2}$$
Question 5
1 / -0

The function $$f : R /{0} \rightarrow R$$ given by $$f(x) =
\dfrac{1}{x} - \dfrac{2}{e^{2x} -1}$$ can be made continuous at $$x=0$$ by
defining $$f(0)$$ as

A
0

B
1

C
2

D
-1

Solution

Given $$f(x)=\dfrac{1}{x}-\dfrac{2}{e^{2x}-1}$$

For $$f(x)$$ to be continuous at $$x=0$$ , its limit should exists at $$x=0$$ and must be finite.

Hence
$$f(0)=\lim\limits_{x\to0}\dfrac{e^{2x}-1-2x}{x(e^{2x}-1)}$$

$$=\lim\limits_{x\to0}\dfrac{1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}.....-1-2x}{x+2x^2+\frac{4x^3}{2!}.....-x}=\dfrac{\frac{4x^2}{2!}}{2x^2}=1$$......[neglecting higher power of $$x$$]
Question 6
1 / -0

$$\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}}$$ is equal to

A
$$e^1$$

B
$$e^{-2}$$

C
$$e^2$$

D
$$e^{-1}$$
Question 7
1 / -0

$$\mathop {\lim }\limits_{x \to 0} {{(1 - \cos 2x)(3 + \cos x)} \over {x\tan 4x}}$$ is equal to

A
2

B
1/2

C
4

D
3

Solution

$$\lim _{ x\rightarrow 0 }{ \cfrac { (1-\cos { 2x } )(3+\cos { x } ) }{ x\tan { 4x } } } \\ \Rightarrow \cfrac { (1-1+2\sin { ^{ 2 }{ x } } )(3+\cos { x } ) }{ x\times 4x } \times (\cfrac { 4x }{ \tan { 4x } } )\\ \Rightarrow \cfrac { 2\sin { ^{ 2 }{ x } } }{ 4{ x }^{ 2 } } \times (3+\cos { x } )\times 1\\ \Rightarrow \cfrac { 1 }{ 2 } { (\cfrac { \sin { x } }{ x } ) }^{ 2 }\times (3+\cos { x } )\times 1\\ \Rightarrow \cfrac { 1 }{ 2 } \times 4\times 1=2$$
Question 8
1 / -0

Let $${P_n} = \prod\limits_{k = 2}^n {\left( {1 - {1 \over {{}^{{}^{k + 1}}{C_2}}}} \right)} .$$ If $$\mathop {\lim }\limits_{x \to \infty } {P_n}$$ can be expressed as lowest rational in the form $$\dfrac { a }{ b } $$ , then value of $$(a+b)$$ is __________.

A
4

B
8

C
10

D
12

Solution
Question 9
1 / -0

$$\lim\limits_{x\to 0}\dfrac{e^{\sin x}-1}{x}=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
none of these

Solution

Now,
$$\lim\limits_{x\to 0}\dfrac{e^{\sin x}-1}{x}$$
$$=\lim\limits_{x\to 0}\dfrac{\dfrac{e^{\sin x}-1}{\sin x}}{\dfrac{x}{\sin x}}$$
$$=\dfrac{\lim\limits_{\sin x\to 0}\dfrac{e^{\sin x}-1}{\sin x}}{\dfrac{1}{\lim\limits_{x\to 0}\dfrac{\sin x}{ x}}}$$
$$=\dfrac{1}{1}$$
$$=1$$.
Question 10
1 / -0

$$\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x = $$

A
$$\infty $$

B
$$ - \infty $$

C
$$0$$

D
does not exist

Solution

$$L=\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x$$
$$\Rightarrow$$ $$L=\tan\dfrac{\pi}{2}$$ [ Applying limit ]
We know that value of $$\tan\dfrac{\pi}{2}$$ is does not exist.
$$\therefore$$ $$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x=$$ does not exist

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Limits and Continuity Test 19

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Limits and Continuity Test 19

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SHARING IS CARING

Question 1

$$(4/23)$$

$$(2/23)$$

$$(1/8)$$

$$(23/4)$$

Question 2

$$1$$

$$-1$$

Does not exist

$$0$$

Solution

Question 3

$$2$$

$$2+e$$

$$2+\dfrac { 1 }{ e }$$

$$3$$

Solution

Question 4

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

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

$$2$$

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

Solution

Question 5

0

1

2

-1

Solution

Question 6

$$e^1$$

$$e^{-2}$$

$$e^2$$

$$e^{-1}$$

Question 7

2

1/2

4

3

Solution

Question 8

4

8

10

12

Solution

Question 9

$$0$$

$$1$$

$$-1$$

none of these

Solution

Question 10

$$\infty $$

$$ - \infty $$

$$0$$

does not exist

Solution

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