Limits and Continuity Test 45

Result

Limits and Continuity Test 45

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

$$\lim _ { x \rightarrow 0 } \frac { \sin x \sin \left( \frac { \pi } { 3 } + x \right) \sin \left( \frac { \pi } { 3 } - x \right) } { x } =$$

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

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

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

D
$$0$$

Solution
Question 2
1 / -0

$$lim_{x\to \dfrac{\pi}{2}} tan^2x(\sqrt{2sin^2x + 3 sin x +4} - \sqrt{sin^2x + 6 sin x+2})$$ is equal to

A
$$\dfrac{3}{4}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{1}{12}$$

D
$$\dfrac{5}{12}$$

Solution
Question 3
1 / -0

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

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$\frac { 1 } { 2 }$$
Question 4
1 / -0

$$P= \lim_{x \rightarrow o^{+}} (1+ \tan^{2} \sqrt{x})^{1/2x}=$$____

A
$$P=1/4$$

B
$$P=1/2$$

C
$$P=2$$

D
$$P=1$$

Solution
Question 5
1 / -0

$$\displaystyle \lim _{ x\rightarrow 0 } \left(\dfrac{1^{1/x}+2^{1/x}+3^{1/x}+.....n^{1/x}}{n}\right)^{nx} ,\ n\ \epsilon \ N $$ is equal to

A
$$n!$$

B
$$1$$

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

D
$$0$$

Solution

$$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { { 1 }^{ 1/x }+{ 2 }^{ 1/x }+{ 3 }^{ 1/x }......+{ n }^{ 1/x } }{ n } \right) }^{ nx }$$
if we put the limits then we have $${ 1 }^{ \infty }$$ form. We have for $$\underset { x\rightarrow \infty }{ lim } { \left[ f\left( x \right) \right] }^{ 9\left( x \right) }$$ such that limit has the form $${ 1 }^{ \infty }$$, then
$$\underset { x\rightarrow \infty }{ lim } { \left[ f\left( x \right) \right] }^{ 9\left( x \right) }={ e }^{ \underset { x\rightarrow \infty }{ lim } 9\left( x \right) \left[ f\left( x \right) -1 \right] }\quad \longrightarrow \left( 1 \right) $$
Using this we can evalute the limit asked,
$$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { { 1 }^{ 1/x }+{ 2 }^{ 1/x }+{ 3 }^{ 1/x }+......{ n }^{ 1/x } }{ n } \right) }^{ nx }={ e }^{ \underset { x\rightarrow \infty }{ lim } nx\left( \dfrac { { 1 }^{ 1/x }+{ 2 }^{ 1/x }+{ 3 }^{ 1/x }......{ n }^{ 1/x }-n }{ n } \right) }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lim } \left( \dfrac { { 1 }^{ 1/x }+{ 2 }^{ 1/x }+{ 3 }^{ 1/x }+......{ n }^{ 1/x }-n }{ \left( 1/x \right) } \right) }$$
if we put the limits we find that it is in $$\div $$ form, so we will employ the L'Hospital's rule
$$={ e }^{ \underset { x\rightarrow \infty }{ lim } \left( \dfrac { { 1 }^{ 1/x }\left( -\dfrac { 1 }{ { x }^{ 2 } } \right) ln1+{ 2 }^{ 1/x }\left( -\dfrac { 1 }{ { x }^{ 2 } } \right) ln2......+{ n }^{ 1/x }\left( -\dfrac { 1 }{ { x }^{ 2 } } \right) lnn-0 }{ \left( -1/{ x }^{ 2 } \right) } \right) }$$
$$={ e }^{ \underset { x\rightarrow \infty }{ lim } \left( { 1 }^{ 1/x }ln1+{ 2 }^{ 1/x }ln2......+{ n }^{ 1/x }lnn \right) }$$
$$={ e }^{ ln1+ln2+ln3......+lnn }$$ [$$\because$$ $$\underset { x\rightarrow \infty }{ lim } { n }^{ 1/x }={ n }^{ 0 }=1$$]
$$={ e }^{ ln\left( 1.2.3.4.5.6......n \right) }$$ [$$\because$$ $$lna+lnb=lnab$$]
$$={ e }^{ lnn! }$$
$$=n!$$ [$$\because$$ $${ e }^{ lnb }=b$$]
Answer : Option A
Question 6
1 / -0

The values of $$\displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt[2]{n^3+1}}$$ is?

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$\infty$$

Solution
Question 7
1 / -0

Let $$f(x)=\displaystyle\lim _{ n\rightarrow \infty }{ \dfrac { { 2x }^{ 2n }\sin { \frac { 1 }{ x } +x } }{ 1+{ x }^{ 2x } } } $$ then find

A
$$\displaystyle\lim _{ x\rightarrow \infty }{ xf\left( x \right) } $$

B
$$\displaystyle\lim _{ x\rightarrow 1 }{ f\left( x \right) } $$

C
$$\displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right) } $$

D
$$\displaystyle\lim _{ x\rightarrow -\infty }{ f\left( x \right) } $$

Solution
Question 8
1 / -0

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

A
$$\dfrac{1}{4}$$

B
$$1$$

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

D
$$None of these$$

Solution
Question 9
1 / -0

Evaluate $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \sin { \left[ \cos { x } \right] } }{ 1+\left[ \cos { x } \right] } } $$ ($$[.]$$ denotes the greatest integer function)

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$does\ not\ exist$$

Solution
Question 10
1 / -0

$$if\left( x \right) =\left[ x-3 \right] +\left[ x-4 \right] \quad for\quad x\epsilon R\quad then\quad \underset { x\rightarrow 3 }{ lim } f\left( x \right) =$$

A
-2

B
-1

C
0

D
2

Solution