Limits and Continuity Test 4

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Limits and Continuity Test 4

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Question 1
1 / -0

lf $$f(x)=2x-3,a=2,l=1$$ and $$\epsilon =0.001$$ then $$\delta>0$$ satisfying$$ 0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon$$, is:

A
$$0.005$$

B
$$0.0005$$

C
$$0.001$$

D
$$0.0001$$

Solution

$$|f(x)-l|<0.001=\epsilon$$
$$\Rightarrow |2x-3-1| < 0.001$$
$$\Rightarrow -0.001<2x-4<0.001$$
$$\Rightarrow -0.0005<x-2<0.0005$$
$$\Rightarrow |x-2|<0.0005$$
$$\Rightarrow |x-a|<0.0005=\delta$$
Hence, $$\delta = 0.0005 > 0$$
Question 2
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 3
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 4
1 / -0

lf $$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$ is finite then the value of $$a,b$$ respectively are

A
$$5\, -4$$

B
$$-5,\ -4$$

C
$$-4,\ 3$$

D
$$4,\ 5$$

Solution

$$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$

As $${ x\rightarrow 0 }$$, denominator tends to 0, so the numerator also tends to 0.

$$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$$
at $$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$$
$$\Rightarrow 1+a+b=0$$
$$\Rightarrow a+b=-1$$
Now, put $$a = -4 $$
$$ \Rightarrow b = 3$$
Option C satisfies above equation.
Question 5
1 / -0

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

A
$$\pi$$

B
$$ 2\pi$$

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

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

Solution

$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$$
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) } $$
It is of the form $$\displaystyle \dfrac{0}{0}$$, so applying L-Hospital's rule
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } } $$
$$=\displaystyle \dfrac{2}{\pi}$$
Question 6
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 7
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 8
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 9
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 10
1 / -0

The value of $$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } $$ is:

A
$$0$$

B
$$1$$

C
$$\displaystyle \frac { \sin { \beta } }{ \beta } $$

D
$$\displaystyle \frac { \sin { 2\beta } }{ 2\beta } $$

Solution

$$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] } $$

$$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] } $$

$$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right) $$ ....... $$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$$

$$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$$ ..... $$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$$
$$=\dfrac { \sin { 2\beta } }{ 2\beta } $$

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Limits and Continuity Test 4

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Limits and Continuity Test 4

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SHARING IS CARING

Question 1

$$0.005$$

$$0.0005$$

$$0.001$$

$$0.0001$$

Solution

Question 2

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

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

$$1$$

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

Solution

Question 3

$$2$$

$$-2$$

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

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

Solution

Question 4

$$5\, -4$$

$$-5,\ -4$$

$$-4,\ 3$$

$$4,\ 5$$

Solution

Question 5

$$\pi$$

$$ 2\pi$$

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

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

Solution

Question 6

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

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

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

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

Solution

Question 7

1

0

-1

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

Solution

Question 8

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

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

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

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

Solution

Question 9

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

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

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

$$0$$

Solution

Question 10

$$0$$

$$1$$

$$\displaystyle \frac { \sin { \beta } }{ \beta } $$

$$\displaystyle \frac { \sin { 2\beta } }{ 2\beta } $$

Solution

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