Limits and Continuity Test 5

Result

Limits and Continuity Test 5

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

Evaluate: $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$

A
$$\displaystyle \frac{3}{2}$$

B
$$\displaystyle \frac{2}{3}$$

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

D
$$\displaystyle \frac{3}{4}$$

Solution

Given,

$$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$

$$\sec x=\dfrac{1}{\cos x}$$

$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$$

$$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

$$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

$$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

as $$\cos 0=1$$ and $$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$

$$=\dfrac{3x}{2x}=\dfrac{3}{2}$$
Question 2
1 / -0

$$\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}$$=

A
$$1$$

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

C
$$\displaystyle \frac{1}{2}$$

D
$$\displaystyle \frac{1}{5}$$

Solution

$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x } $$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } } $$
$$\displaystyle =\dfrac{2}{\pi}$$
Question 3
1 / -0

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$=

A
$$\sqrt{2}$$

B
$$\displaystyle \frac{1}{\sqrt{2}}$$

C
$$-\sqrt{2}$$

D
$$\displaystyle \frac{-1}{\sqrt{2}}$$

Solution

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }$$
$$ =\sqrt { 2 } $$
Question 4
1 / -0

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$$

A
$$\sqrt{3}$$

B
$$\dfrac{1}{\sqrt{3}}$$

C
$$-\sqrt{3}$$

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

Solution

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }$$
$$=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }$$
$$=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}$$
Question 5
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$$=

A
$$\displaystyle \log(\frac{a}{b})$$

B
$$\displaystyle \log(\frac{b}{a})$$

C
$$\log(ab)$$

D
$${\log_{b}}{a}$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$$

$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-1-b^{x}+1}{{x}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(a^{x}-1)-(b^{x}-1)}{{x}}$$
$$=\log a - \log b$$ ......... Using formula
$$=\log\left(\dfrac{a}{b}\right)$$
Hence, $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}=\log\left(\dfrac{a}{b}\right)$$
Question 6
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=$$

A
0

B
1

C
$$\displaystyle \frac{\pi}{180}$$

D
$$\pi$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$$
$$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x } $$
$$\displaystyle={ \dfrac { \pi }{ 180 } }$$
Question 7
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 8
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 9
1 / -0

$$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$$

A
$$1$$

B
$$-1$$

C
$$-\displaystyle \frac{1}{2}$$

D
$$\displaystyle \frac{1}{2}$$

Solution

$$Put\ \theta = \dfrac{\pi }{2}-\phi $$

$$\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}$$
Question 10
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.
$$0\times finite\ no.=0$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 5

Result

Limits and Continuity Test 5

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \frac{3}{2}$$

$$\displaystyle \frac{2}{3}$$

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

$$\displaystyle \frac{3}{4}$$

Solution

Question 2

$$1$$

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

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{1}{5}$$

Solution

Question 3

$$\sqrt{2}$$

$$\displaystyle \frac{1}{\sqrt{2}}$$

$$-\sqrt{2}$$

$$\displaystyle \frac{-1}{\sqrt{2}}$$

Solution

Question 4

$$\sqrt{3}$$

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

$$-\sqrt{3}$$

$$-\dfrac{1}{\sqrt{3}}$$

Solution

Question 5

$$\displaystyle \log(\frac{a}{b})$$

$$\displaystyle \log(\frac{b}{a})$$

$$\log(ab)$$

$${\log_{b}}{a}$$

Solution

Question 6

0

1

$$\displaystyle \frac{\pi}{180}$$

$$\pi$$

Solution

Question 7

$$0$$

$$1$$

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

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

Solution

Question 8

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

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

$$4$$

$$-4$$

Solution

Question 9

$$1$$

$$-1$$

$$-\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{1}{2}$$

Solution

Question 10

does not exist

is equal to 0

is equal to 1

exists and different from 0 and 1

Solution

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.