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

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SHARING IS CARING

Question 1

−12-\dfrac{1}{2}−21​

12\dfrac{1}{2}21​

111

32\dfrac{3}{2}23​

Solution

Question 2

111

2π\displaystyle \frac{2}{\pi}π2​

12\displaystyle \frac{1}{2}21​

15\displaystyle \frac{1}{5}51​

Solution

Question 3

12\displaystyle \frac{1}{2}21​

32\displaystyle \frac{3}{2}23​

34\displaystyle \frac{3}{4}43​

14\displaystyle \frac{1}{4}41​

Solution

Question 4

34\displaystyle \frac{3}{4}43​

14\displaystyle \frac{1}{4}41​

43\displaystyle \frac{4}{3}34​

000

Solution

Question 5

103\dfrac{10}{3}310​

310\dfrac{3}{10}103​

65\dfrac{6}{5}56​

56\dfrac{5}{6}65​

Solution

Question 6

111

000

−1-1−1

12\displaystyle \dfrac{1}{2}21​

Solution

Question 7

2\sqrt{2}2​

12\displaystyle \frac{1}{\sqrt{2}}2​1​

−2-\sqrt{2}−2​

−12\displaystyle \frac{-1}{\sqrt{2}}2​−1​

Solution

Question 8

222

−2-2−2

12\displaystyle \frac{1}{2}21​

−12-\displaystyle \frac{1}{2}−21​

Solution

Question 9

3\sqrt{3}3​

13\dfrac{1}{\sqrt{3}}3​1​

−3-\sqrt{3}−3​

−13-\dfrac{1}{\sqrt{3}}−3​1​

Solution

Question 10

0

1

π180\displaystyle \frac{\pi}{180}180π​

π\piπ

Solution

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$-\dfrac{1}{2}$

$\dfrac{1}{2}$

$1$

$\dfrac{3}{2}$

$1$

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

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

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

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

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

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

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

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

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

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

$0$

$\dfrac{10}{3}$

$\dfrac{3}{10}$

$\dfrac{6}{5}$

$\dfrac{5}{6}$

$1$

$0$

$-1$

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

$\sqrt{2}$

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

$-\sqrt{2}$

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

$2$

$-2$

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

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

$\sqrt{3}$

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

$-\sqrt{3}$

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

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

$\pi$