Limits and Derivatives Test

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Limits and Derivatives Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$=

A
$$0$$

B
$$1$$

C
$$(\displaystyle \frac{\pi}{180})^{3}$$

D
$$4.(\displaystyle \frac{\pi}{180})^{3}$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } } $$
$$\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } } $$
$$\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }$$
Question 2
1 / -0

Solve:
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}$$

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

B
$$\dfrac{3}{4}$$

C
$$4$$

D
$$-4$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3\tan x-\tan 3x}{2x^{3}}$$

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+......)-(3x+\dfrac{(3x)^3}{3}+\dfrac{2(3x)^5}{15}+....)}{2x^{3}}=\dfrac{1-9}{2}=-4$$
Question 3
1 / -0

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$$=

A
1

B
0

C
-1

D
$$\displaystyle \dfrac{1}{\pi}$$

Solution

$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } } $$

$$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }$$

$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } } $$

$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1$$
Question 4
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})$$ is:

A
does not exist

B
is equal to 0

C
is equal to 1

D
exists and different from 0 and 1

Solution

$$sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)$$
$$sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$$
$$\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)$$
sin$$(\dfrac{1}{x})$$ is an oscillatory function means same finite no.
$$0\times finite\ no.=0$$
Question 5
1 / -0

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$=

A
$$2$$

B
$$1$$

C
$$0$$

D
$$3$$

Solution

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}$$.................(dividing the num and deno by cosx)
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1$$
Question 6
1 / -0

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$=

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

B
$$\displaystyle \frac{1}{4}$$

C
$$\displaystyle \frac{1}{6}$$

D
$$\displaystyle \frac{1}{8}$$

Solution

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } } $$......(x-->h-pi/2)
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }$$

$$\displaystyle =\dfrac { 1 }{ 8 } $$
Question 7
1 / -0

$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$$

A
$$1$$

B
$$-1$$

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

D
$$0$$

Solution

$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}$$

$$=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}$$

$$=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1$$
Question 8
1 / -0

lf $$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } } $$,then $$\displaystyle \lim _{ x\rightarrow \infty } f(x)$$=

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

B
$$-1$$

C
$$0$$

D
$$1$$

Solution

$$\displaystyle \lim _{ x\rightarrow \infty } f(x)$$$$\displaystyle = \lim _{ x\rightarrow \infty }\sqrt { \dfrac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } }=\lim _{ x\rightarrow \infty }\sqrt { \dfrac { 1-\dfrac{\sin^2x}{x}}{ 1+\dfrac{\cos x}{x} }}=\sqrt{\dfrac{1-0}{1-0}}=1$$
Question 9
1 / -0

lf $$ \mathrm{f}(\mathrm{x})=0$$ has a repeated root $$ \alpha$$, then another equation having $$\alpha$$ as root, is

A
$$\mathrm{f}(2\mathrm{x})=0$$

B
$$\mathrm{f}(3\mathrm{x})=0$$

C
$$\mathrm{f}^{'}(\mathrm{x})=0$$

D
$$\mathrm{f}^{''}(\mathrm{x})=0$$

Solution

Repeated roots mean that the curve will touch the $$x$$-axis.
The curve will have a convex or concave part at the repeated root $$\alpha$$ or a point of inflection.
In any case $$f'(x)$$ will always be zero at the point of repeated root $$\alpha$$.
Question 10
1 / -0

$$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$$

A
$$\displaystyle \pi$$

B
$$\displaystyle \frac{\pi}{2}$$

C
$$\dfrac{\pi}{8}$$

D
$$\displaystyle \frac{\pi}{4}$$

Solution

As $$ x \rightarrow \infty , \cos \left (\dfrac{\pi}{8x}\right) \rightarrow 1 $$
Looking at the rest of the part,
$$\underset{ x \rightarrow \infty}{\lim} \dfrac{\pi}{8x} \rightarrow 0 $$
$$\underset{ x \rightarrow \infty}{\lim} x.\sin (\dfrac{\pi}{8x}) $$
$$=\underset{ x \rightarrow \infty}{\lim} \dfrac{\sin (\dfrac{\pi}{8x})}{\dfrac{1}{x}} $$
We multiply and divide by $$ \dfrac{\pi}{8} $$
$$=\underset{ x \rightarrow \infty}{\lim} , \dfrac{\sin \left (\dfrac{\pi}{8x}\right)}{\dfrac{\pi}{8x}} \times \dfrac{\pi}{8x} $$
Now using the property of a limit,
$$\underset{ y \rightarrow 0}{\lim} , \dfrac{\sin y}{y} $$ $$= 1$$
We get, applying the limit,
$$= 1 \times $$ $$ \dfrac{\pi}{8} $$
$$=$$ $$ \dfrac{\pi}{8} $$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 14

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Limits and Derivatives Test - 14

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$(\displaystyle \frac{\pi}{180})^{3}$$

$$4.(\displaystyle \frac{\pi}{180})^{3}$$

Solution

Question 2

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

$$\dfrac{3}{4}$$

$$4$$

$$-4$$

Solution

Question 3

1

0

-1

$$\displaystyle \dfrac{1}{\pi}$$

Solution

Question 4

does not exist

is equal to 0

is equal to 1

exists and different from 0 and 1

Solution

Question 5

$$2$$

$$1$$

$$0$$

$$3$$

Solution

Question 6

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

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{1}{8}$$

Solution

Question 7

$$1$$

$$-1$$

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

$$0$$

Solution

Question 8

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

$$-1$$

$$0$$

$$1$$

Solution

Question 9

$$\mathrm{f}(2\mathrm{x})=0$$

$$\mathrm{f}(3\mathrm{x})=0$$

$$\mathrm{f}^{'}(\mathrm{x})=0$$

$$\mathrm{f}^{''}(\mathrm{x})=0$$

Solution

Question 10

$$\displaystyle \pi$$

$$\displaystyle \frac{\pi}{2}$$

$$\dfrac{\pi}{8}$$

$$\displaystyle \frac{\pi}{4}$$

Solution

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