Limits and Derivatives Test

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Limits and Derivatives Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$$

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

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

C
$$1$$

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

Solution

We simplify the given expression as,

$$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)} $$

Let $$x - sin x = y$$

As $$ x \rightarrow 0 \text{ so does y } \rightarrow 0 $$

Hence, the question transforms into

$$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y}) $$

$$= \dfrac{1}{2} \times 1 $$

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

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}$$=

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

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

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

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

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}$$

$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$$...... As$$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$$
Question 4
1 / -0

Solve : $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$

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

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

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

D
$$0$$

Solution

Solve : $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$
$$=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)$$
$$=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}$$
$$=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}$$
$$=\dfrac34$$
Question 5
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=$$

A
$$\dfrac{10}{3}$$

B
$$\dfrac{3}{10}$$

C
$$\dfrac{6}{5}$$

D
$$\dfrac{5}{6}$$

Solution

Using, $$1 - cos 2x = 2{sin}^{2} x $$,
The expression transforms to,
$$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x} $$
Rewriting the expression in a different form,
$$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3} $$
Therefore, the limit to the expression is $$ \dfrac{10}{3} $$
Hence, option 'A' is correct.
Question 6
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$$

A
$$1$$

B
$$0$$

C
$$-1$$

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

Solution

$$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$$

$$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$$

$$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) } $$

$$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$$
Question 7
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 8
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$=

A
$$2$$

B
$$-2$$

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

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

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$

$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x } }{ 1-\tan ^{ 2 }{ x } } -2x\tan x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x } }{ 1+\tan ^{ 2 }{ x } } \right)^{ 2 } } } $$

$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}$$

$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } } $$

$$=\displaystyle \dfrac { 1 }{ 2 } $$
Question 9
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 10
1 / -0

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

A
0

B
1

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

D
$$\pi$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$$
$$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x } $$
$$\displaystyle={ \dfrac { \pi }{ 180 } }$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 13

Result

Limits and Derivatives Test - 13

Score

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Rank

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SHARING IS CARING

Question 1

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

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

$$1$$

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

Solution

Question 2

$$1$$

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

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

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

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

$$0$$

Solution

Question 5

$$\dfrac{10}{3}$$

$$\dfrac{3}{10}$$

$$\dfrac{6}{5}$$

$$\dfrac{5}{6}$$

Solution

Question 6

$$1$$

$$0$$

$$-1$$

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

Solution

Question 7

$$\sqrt{2}$$

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

$$-\sqrt{2}$$

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

Solution

Question 8

$$2$$

$$-2$$

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

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

Solution

Question 9

$$\sqrt{3}$$

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

$$-\sqrt{3}$$

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

Solution

Question 10

0

1

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

$$\pi$$

Solution

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