Limits and Continuity Test 20

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Limits and Continuity Test 20

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Question 1
1 / -0

Suppose the function $$f(x)-f(2x)$$ has the derivative $$5$$ at $$x=1$$ and derivative $$7$$ at $$x=2$$. The derivative of the function $$f(x)-f(4x)$$ at $$x=1$$, has the value equal to?

A
$$19$$

B
$$9$$

C
$$17$$

D
$$14$$

Solution
Question 2
1 / -0

$$\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x } \right) }{ \left( \tan ^{ 2 }{ 2x } \right) = } }$$

A
$$1$$

B
$$2$$

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

D
$$Does\ not\ exist$$

Solution

$$\mathop {\lim }\limits_{x \to 0} {\log _{{{\tan }^2}x}}\left( {{{\tan }^2}2x} \right)$$

Let,

$$\begin{array}{l} y={ \log _{ { { \tan }^{ 2 } }x } }{ \tan ^{ 2 } }2x \\ =\log { \tan ^{ 2 } } 2x={ \left( { { { \tan }^{ 2 } }x } \right) ^{ y } } \\ =2\log \tan 2x=2y\log \tan x \\ \Rightarrow y=\dfrac { { \log \tan 2x } }{ { \log \tan x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \log \tan 2x } }{ { \log \tan x } } \end{array}$$

Using L-Hospital Rule

$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \dfrac { 1 }{ { \tan 2x } } \times { { \sec }^{ 2 } }2x\times 2 } }{ { \dfrac { 1 }{ { \tan x } } { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2{ { \sec }^{ 2 } }2x } }{ { \tan 2x } } \times \dfrac { { \tan x } }{ { { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\tan x } }{ { \tan 2x } } \, \, \, \, \, \, \left( { { \lim }_{ x\to 0 }\sec x=4 } \right) \end{array}$$

$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\dfrac { { \tan x } }{ x } \times x } }{ { \dfrac { { \tan 2x } }{ { 2x } } \times 2x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\, \, \, 1\, \, \left( { { \lim }_{ x\to 0 }\dfrac { { \tan x } }{ x } =1 } \right) \\ \Rightarrow { \lim }_{ x\to 0 }{ \log _{ { { \tan }^{ 2 } }x } }\left( { { { \tan }^{ 2 } }2x } \right) =1 \end{array}$$
Question 3
1 / -0

$$\displaystyle \lim_{n \rightarrow \infty} {^{n}C_{c}}\left(\dfrac {m}{n}\right)^{x}\left(1-\dfrac {m}{n}\right)^{n-x}$$ equal to

A
$$\dfrac {M^{x}}{x\ !}.e^{-m}$$

B
$$\dfrac {M^{x}}{x\ !}.e^{m}$$

C
$$0$$

D
$$\dfrac {m^{x+1}}{me^{m}x\ !}$$

Solution

Given $$\displaystyle\lim_{n\rightarrow \infty}{^{n}}C_x \bigg(\dfrac{m}{n}\bigg)^{x}\bigg(1-\dfrac{m}{n}\bigg)^{n-x}=$$coefficient of $$x$$ in $$\displaystyle\lim_{n\rightarrow\infty}\bigg(\dfrac{m}{n}+1-\dfrac{m}{n}\bigg)^{n}$$
$$=$$coeeficient of $$x$$ in $$\displaystyle\lim_{n\rightarrow\infty}\bigg(1\bigg)^{n}=$$coefficient of $$x$$ in $$1=0$$
Question 4
1 / -0

If [.] denotes, GIF , then $$\underset{x \rightarrow 0}{lt} \left( \left[\dfrac{2018 sin^{-1} x}{x}\right] + \left[\dfrac{2020x}{tan^{-1} x}\right]\right)$$ =

A
4038

B
4037

C
4036

D
4039

Solution
Question 5
1 / -0

The value of $$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$$

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

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

C
$$2$$

D
none of these

Solution

Now,
$$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$$
$$=\lim\limits_{x\to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x^2}$$
$$=\dfrac{1}{2}\times\left(\lim\limits_{\dfrac{x}{2}\to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2$$
$$=\dfrac{1}{2}\times 1$$
$$=\dfrac{1}{2}$$.
Question 6
1 / -0

$$\displaystyle\mathop {\lim }\limits_{\ x \to 0} \cos \frac{x}{2}\cos \frac{x}{{{2^2}}}\cos \frac{x}{{{2^3}}}......\cos \frac{x}{{{2^n}}}$$ is equal to

A
$$1$$

B
$$-1$$

C
$$\dfrac{{\sin x}}{x}$$

D
$$\dfrac{x}{{\sin \,x}}$$

Solution

$$\displaystyle\lim_{x\rightarrow \infty}\left(\cos\dfrac{x}{2}\cos \dfrac{x}{2^{2}}, \cos\dfrac{x}{2^{3}}....... \cos\dfrac{x}{2^{x}}\right)$$

$$\sin 2\theta =2\sin \theta\cos\theta$$ $$2\theta=x$$

$$=\sin x=2\left(\sin\dfrac{x}{2}\right)\cos\dfrac{x}{2}$$ $$\theta=\dfrac{x}{2}$$

$$=2\left(2\sin \dfrac{x}{4}\cos \dfrac{x}{4}\right)\cos \dfrac{x}{2}$$ $$2\theta=\dfrac{x}{2}$$

$$\theta=\dfrac{x}{4}$$

$$=8\left(\left(\sin\dfrac{x}{8}\right)\cos\dfrac{x}{8}\cos\dfrac{x}{4}\cos\dfrac{x}{2}\right)$$ $$\dfrac{x}{4}=2\theta$$

$$\theta=\dfrac{x}{8}$$

$$=16\left(\sin \dfrac{x}{16}\cos\dfrac{x}{4}\cos\dfrac{x}{8}\cos\dfrac{x}{4}\cos\dfrac{x}{2}\right)\theta=\dfrac{x}{16}$$

$$=16\left(\sin\dfrac{x}{2^{4}}\cos\dfrac{x}{2^{4}}\cos\dfrac{x}{2^{3}}\cos \dfrac{x}{2^{2}}\cos\dfrac{x}{2}\right)$$

$$\sin x=\left(2^{n}\sin\dfrac{x}{2^{x}}\right)\cos\dfrac{x}{2x}\cos\dfrac{x}{2^{x-1}}.......\cos\dfrac{x}{2^{3}}\cos\dfrac{x}{2^{2}}\cos\dfrac{x}{2}$$

$$\left.\dfrac{\sin x}{2^{x}\sin\dfrac{x}{2^{x}}}\right]=\cos \dfrac{x}{2}\cos\dfrac{x}{2^{2}}\cos\dfrac{x}{2^{3}}..... \cos\dfrac{x}{2^{x}}$$

$$\displaystyle\lim_{x\rightarrow \infty}\dfrac{\sin x}{2^{n}\sin\left(\dfrac{x}{2^{x}}\right)}\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$

$$\overline { \infty } =0$$

$$\displaystyle\lim_{x\rightarrow \infty}\dfrac{\sin x}{2^{x}\dfrac{\sin\left(\dfrac{x}{2^{x}}\right)}{\left(\dfrac{x}{2^{x}}\right)}\times \left(\dfrac{x}{2^{x}}\right)}=\left.\dfrac{\sin x}{x}\right]\lim_{x\rightarrow \infty}\dfrac{\dfrac{1}{\sin\left(\dfrac{x}{2^{x}}\right)}}{\dfrac{x}{2^{x}}}$$

$$=\dfrac{\sin x}{x}$$
Question 7
1 / -0

$$\underset{h \rightarrow 0}{lim} \dfrac{\sqrt{x + h} -\sqrt{x}}{h}$$ is equal to

A
$$\sqrt{x}$$

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

C
$$2 \sqrt{x}$$

D
$$\dfrac{1}{\sqrt{x}}$$

Solution

The given limit can be written as,

$$\underset { h\rightarrow 0 }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ h } =\underset { x+h\rightarrow x }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ (x+h)-x } $$

We know that,
$$\underset { x\rightarrow a }{ lim } \dfrac { { x }^{ n }-{ a }^{ n } }{ x-a } =n{ a }^{ n-1 }$$

So, the given limit is,
$$=\dfrac { 1 }{ 2 } { x }^{ \left( \dfrac { 1 }{ 2 } -1 \right) }=\dfrac { 1 }{ 2\sqrt { x } } $$
Question 8
1 / -0

Evaluate:
$$\lim\limits_{x\to 0}(1+ax)^{\dfrac{1}{x}}$$

A
$$e^a$$

B
$$e^{\dfrac{1}{a}}$$

C
$$1$$

D
None of these

Solution

Now,
$$\lim\limits_{x\to 0}(1+ax)^{\dfrac{1}{x}}$$
$$=\left(\lim\limits_{ax\to 0}(1+ax)^{\dfrac{1}{ax}}\right)^a$$
$$=e^a$$.
Question 9
1 / -0

$$\lim\limits_{x\to 0}\dfrac{1-\cos x }{x^2}=$$

A
$$4$$

B
$$2$$

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

D
$$1$$

Solution

Now,
$$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$$
$$=\lim\limits_{x\to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x^2}$$
$$=\dfrac{1}{2}\lim\limits_{x\to 0}\dfrac{\sin^2\dfrac{x}{2}}{\dfrac{x^2}{4}}$$
$$=\dfrac{1}{2}\left(\lim\limits_{\dfrac{x}{2}\to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2$$
$$=\dfrac{1}{2}.1$$
$$=\dfrac{1}{2}$$.
Question 10
1 / -0

$$\underset{x \to 0}{\lim}\dfrac{\sin [ \cos x]}{1+[\cos x]}$$ is

A
$$1$$

B
$$0$$

C
does not exist

D
$$2$$

Solution

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 20

Result

Limits and Continuity Test 20

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SHARING IS CARING

Question 1

$$19$$

$$9$$

$$17$$

$$14$$

Solution

Question 2

$$1$$

$$2$$

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

$$Does\ not\ exist$$

Solution

Question 3

$$\dfrac {M^{x}}{x\ !}.e^{-m}$$

$$\dfrac {M^{x}}{x\ !}.e^{m}$$

$$0$$

$$\dfrac {m^{x+1}}{me^{m}x\ !}$$

Solution

Question 4

4038

4037

4036

4039

Solution

Question 5

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

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

$$2$$

none of these

Solution

Question 6

$$1$$

$$-1$$

$$\dfrac{{\sin x}}{x}$$

$$\dfrac{x}{{\sin \,x}}$$

Solution

Question 7

$$\sqrt{x}$$

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

$$2 \sqrt{x}$$

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

Solution

Question 8

$$e^a$$

$$e^{\dfrac{1}{a}}$$

$$1$$

None of these

Solution

Question 9

$$4$$

$$2$$

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

$$1$$

Solution

Question 10

$$1$$

$$0$$

does not exist

$$2$$

Solution

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