Limits and Continuity Test 38

Result

Limits and Continuity Test 38

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

$$ \underset { x\rightarrow 0 }{ lim } \left[ { x }^{ 2 }cosec\quad \left( { x }^{ 2 } \right)^0 \right] $$is equal to :

A
$$\cfrac { \pi} {180} $$

B
$$ \cfrac { \pi} {90} $$

C
$$ {0} $$

D
$$\cfrac {90} { \pi } $$

Solution
Question 2
1 / -0

The value of $$\lim_{x \rightarrow -1} \dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}$$ is given by

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

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

C
$$1$$

D
$$0$$

Solution
Question 3
1 / -0

If $$\lim_{x \rightarrow 0}\dfrac{a \sin x-bx+cx^{2}+x^{3}}{2x^{2} \log(1+x)-2x^{3}+x^{4}}$$ exists and is finite, then the value of $$a,b,c$$ are respectively

A
$$0,6,6$$

B
$$6,0,6$$

C
$$6,6,0$$

D
$$0,0,6$$

Solution

Given, $$\underset{x \rightarrow 0}{\lim} \dfrac{a \sin x - bx + cx^2 + x^3}{2x^2 \log (1 + x) - 2x^3 + x^4} = $$ finite

To find : $$a, b, c$$

$$\underset{x \rightarrow 0}{\lim} \dfrac{a \sin x - bx + cx^2 + x^3}{2x^2 \log (1 + x) - 2x^3 + x^4} \rightarrow \dfrac{0}{0}$$

$$\therefore$$ Using L'hospital rule

$$\underset{x \rightarrow 0}{\lim} \dfrac{a \cos x - b + 2 cx + 3x^2}{4x \ln (1 + x) + \dfrac{2x^2}{(1 + x)} - 6x^2 + 4x^3} \rightarrow \dfrac{a - b}{0}$$

Using L'Hospital again,

$$\underset{x \rightarrow 0}{\lim} \dfrac{-a \sin x + 2c + 6x}{4 \ln (1 + x) + \dfrac{4x}{1 + x} + \dfrac{4x (1 + x) - 2x^2}{(1 + x)^2} - 12x + 12x^2} \rightarrow \dfrac{2c}{0}$$

Using L' Hospital rule again,

$$\underset{x \rightarrow 0}{\lim} \dfrac{-a \cos x + 6}{\dfrac{(4 + 4x)(1 +x)^2 - (2 + 2x)(4x + 2x^2) - 12 + 24x}{(1 + x)^4}}$$ $$+\dfrac{4}{1 + x} + \dfrac{4 (1 + x) - 4x}{(1 + x)^2}$$

$$= \dfrac{-a + 6}{4 + 4 + 4 - 12} = \dfrac{-a + 6}{0} \Rightarrow a = 6$$

as the limit is finite.

Hence, $$a = 6 = b, c = 0$$. Option C.
Question 4
1 / -0

$$ \underset { x\rightarrow a }{ lim } \cfrac { sin\quad x-sin\quad a }{ \sqrt [ 3 ]{ x } -\sqrt [ 3 ]{ a } } $$

A
$$ \sqrt [ 3 ]{ a\quad cos\quad a } $$

B
$$2\sqrt [ 3 ]{ a } $$

C
$$ { 3a }^{ { 2 /}{ 3 } }cos\quad a $$

D
$$ \sqrt [ 2 ]{ a\quad cos\quad a } $$

Solution
Question 5
1 / -0

The value of $$\displaystyle \lim_{x\rightarrow 0}\dfrac {1-\cos^{3}x}{x\sin x\cos x}$$ is

A
$$\dfrac {2}{5}$$

B
$$\dfrac {3}{5}$$

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

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

Solution
Question 6
1 / -0

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

A
$$2$$

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

C
$$-2$$

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

Solution
Question 7
1 / -0

The value of $$\lim_{x \rightarrow 0} \left(\dfrac{\tan x}{x}\right)^{1/x^{3}}$$ is-

A
$$0$$

B
$$\infty$$

C
$$e^{1/4}$$

D
$$Does\ not\ exist$$

Solution
Question 8
1 / -0

$$\displaystyle \lim_{x\rightarrow 0^{+}}{(\csc x)^{1/\log x}}$$=

A
$$e$$

B
$$e^{-1}$$

C
$$e^{2}$$

D
$$1$$

Solution
Question 9
1 / -0

$$ \underset { x\rightarrow \cfrac { \pi }{ 2 } }{ lim } \cfrac { cot \, x-cos\, x }{ \left( \pi -{ 2x } \right)^ 3 } $$ equals

A
$$ \cfrac {1} {24} $$

B
$$ \cfrac {1} {16} $$

C
$$ \cfrac {1} {8} $$

D
$$ \cfrac {1} {4} $$

Solution

Given,

$$\lim _{x\to \frac{\pi }{2}}\left(\dfrac{\cot \left(x\right)-\cos \left(x\right)}{\left(\pi -2x\right)^3}\right)$$

apply L-Hospital's rule

$$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{-\csc ^2\left(x\right)+\sin \left(x\right)}{-6\left(\pi -2x\right)^2}\right)$$

$$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{-\csc ^2\left(x\right)+\sin \left(x\right)}{6\left(\pi -2x\right)^2}\right)$$

apply L-Hospital's rule

$$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{-2\csc ^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{-24\left(\pi -2x\right)}\right)$$

$$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{-2\csc ^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{24\left(\pi -2x\right)}\right)$$

apply L-Hospital's rule

$$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)}{-48}\right)$$

$$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{1}{48}\left(2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)\right)\right)$$

$$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)}{48}\right)$$

$$=-\dfrac{2\left(-2\csc ^2\left(\frac{\pi }{2}\right)\cot ^2\left(\frac{\pi }{2}\right)-\csc ^4\left(\frac{\pi }{2}\right)\right)-\sin \left(\frac{\pi }{2}\right)}{48}$$

upon solving, we get,

$$=\dfrac{1}{16}$$
Question 10
1 / -0

The value of $$ \underset { x\rightarrow \frac { x }{ 2 } }{ lim } \frac { log\sin { x } }{ { \left( \frac { \pi }{ 2 } -x \right) }^{ 2 } }$$ is

A
$$0$$

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

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

D
$$-2$$

Solution