Limits and Derivatives Test

Result

Limits and Derivatives Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

$$\underset { x\rightarrow 1 }{ lim } \frac { xtan(x-[x]) }{ x-1 } $$ is:

A
1

B
0

C
-1

D
does not exist

Solution
Question 2
1 / -0

The value of $$\displaystyle \lim_{x \rightarrow 0} \left(\dfrac{\sin x}{x}\right)^{1/x^{2}}$$ is

A
$$e^{-1/6}$$

B
$$e^{1/6}$$

C
$$e^{-1/3}$$

D
$$e^{1/3}$$

Solution
Question 3
1 / -0

If z = z(x) and $$(2cosx)\frac { dz }{ dx } +(sinx)z=sinx$$, z(0) = 3, then $$z(\frac { \pi }{ 2 } )$$ equals :

A
1

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

C
$$\frac { 5 }{ 2 } $$

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

Solution
Question 4
1 / -0

For the function, $$f(x) = (x - \frac{1}{x})^2$$, the first derivative with respect to x is

A
$$2 (x - \frac{1}{x^3})$$

B
$$2 (x - \frac{1}{x})$$

C
$$2 (x + \frac{1}{x^2})$$

D
$$2 (x - \frac{1}{x^2})$$

Solution
Question 5
1 / -0

The value of $$lim_{x\to 0} \dfrac{sin(\pi cos^2 x)}{x^2}$$ equals

A
$$-\pi$$

B
$$\pi$$

C
$$\dfrac{\pi}{2}$$

D
$$2 \pi$$

Solution
Question 6
1 / -0

The value of $$lim_{\theta \to \dfrac{\pi}{2}} (sec \theta - tan \theta)$$ equals

A
$$0$$

B
$$1$$

C
$$2$$

D
$$\infty$$

Solution
Question 7
1 / -0

$$\lim- {x\to 0}$$ $$\dfrac{1- cos(1 - cos4x)}{x^4}$$ is equal to :

A
4

B
16

C
32

D
None of these

Solution
Question 8
1 / -0

The value of $$\displaystyle\lim_{x\to 0} |x|^{sinx}$$ equals

A
$$0$$

B
$$-1$$

C
$$1$$

D
does not exist

Solution
Question 9
1 / -0

If $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \left( \sin { nx } \right) \left[ (a-n)nx-tanx \right] }{ { x }^{ 2 } } } =0$$, then the value of $$a$$

A
$$\dfrac { 1 }{ n }$$

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

C
$$n+\dfrac{1}{n}$$

D
$$None\ of\ these$$

Solution
Question 10
1 / -0

Arrange the following limits in the ascending order :
(1) $$\lim _ { x \rightarrow \infty } \left( \dfrac { 1 + x } { 2 + x } \right) ^ { x + 2 }$$

(2) $$\lim _ { x \rightarrow 0 } ( 1 + 2 x ) ^ { 3 / x }$$

(3) $$\lim _ { \theta \rightarrow 0 } \dfrac { \sin \theta } { 2 \theta }$$

(4) $$\lim _ { x \rightarrow 0 } \dfrac { \log _ { e } ( 1 + x ) } { x }$$

A
$$1,2,3,4$$

B
$$1,3,4,2$$

C
$$1,4,3,2$$

D
$$3,4,1,2$$

Solution

(i) $$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { 1+x }{ 2+x } \right) }^{ x+2 }$$
$$=$$ $${ \left( \dfrac { \dfrac { 1 }{ x } +1 }{ \dfrac { 2 }{ x } +1 } \right) }^{ x+2 }$$
$$=\underset { x\rightarrow \infty }{ lim } { e }^{ ln{ \left( \dfrac { 1+x }{ 2+x } \right) }^{ 2+x } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \left( 2+x \right) ln\left( \dfrac { 1+x }{ 2+x } \right) }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { ln\left( 1+x \right) -ln\left( 2+x \right) }{ \left( 1/2+x \right) } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \dfrac { 1 }{ 1+x } -\dfrac { 1 }{ 2+x } }{ \dfrac { -1 }{ { \left( 2+x \right) }^{ 2 } } } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { { \left( 2+x \right) }^{ 2 }\left[ \left( 2+x \right) -\left( 1+x \right) \right] }{ \left( 2+x \right) -\left( 1+x \right) \left( 2+x \right) \left( 1+x \right) } }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \left( 2+x \right) }{ 1+x } }$$
$$={ e }^{ 1 }$$
(ii) $$\underset { x\rightarrow 0 }{ lim } { \left( 1+2x \right) }^{ 3/x }$$
$$={ e }^{ \underset { x\rightarrow 0 }{ lt } \left( 2x \right) 3/x }$$
$$={ e }^{ 6 }$$
(iii) $$\underset { \theta \rightarrow 0 }{ lt } \dfrac { \sin\theta }{ 2\theta } =\underset { \theta \rightarrow 0 }{ lt } \dfrac { \cos\theta }{ 2 } =\dfrac { 1 }{ 2 } $$
(iv) $$\underset { x\rightarrow 0 }{ lt } \dfrac { { log }_{ e }\left( 1+x \right) }{ x } .\underset { x\rightarrow 0 }{ lt } \dfrac { 1 }{ 1+x } =1$$
$$\therefore$$ $$3<4<1<2$$.