Limits and Continuity Test 27

Result

Limits and Continuity Test 27

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

$$\underset { x\rightarrow 0 }{ lim } \frac { tan(sinx)-x }{ { tanx }^{ 3 } } $$ is equal to

A
$$\frac { 1 }{ 6 } $$

B
$$\frac { 1 }{ 3 } $$

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

D
1
Question 2
1 / -0

The value of $$\displaystyle n\xrightarrow { lim } \infty\frac{1.n+2.(n-1)+3.(n-2)+...+n.1}{{1}^{2}+{2}^{2}+...+{n}^{2}}$$ is

A
$$1$$

B
$$-1$$

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

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

Solution

Now,
$$1.n+2(n-1)+3.(n-2)+....+n.1$$
The general term of the above series is
$${ T }_{ r }=r(n-r+1)=r(n+1-r)$$
Sum of series$$=S=\sum _{ r=1 }^{ n }{ { T }_{ r } } =\sum _{ r=1 }^{ n }{ r(n+1-r) } $$
$$S=\sum _{ r=1 }^{ n }{ nr } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\sum _{ r=1 }^{ n }{ r } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\cfrac { n(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 } $$
Similarly, $${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ n }^{ 2 }=\cfrac { n(n+1)(2n+1) }{ 6 } $$
$$\lim _{ n\rightarrow \infty }{ \cfrac { 1.n+2(n-1)+3.(n-2)+...+n.1 }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+........{ n }^{ 2 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { { n }^{ 2 }(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 } }{ \cfrac { n(n+1)(2n+1) }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { n }{ 2 } +\cfrac { 1 }{ 2 } -\cfrac { 2n+1 }{ 6 } }{ \cfrac { 2n+1 }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ (\cfrac { 3n+3-2n-1 }{ 2n+1 } ) } =\lim _{ n\rightarrow \infty }{ (\cfrac { 3+\cfrac { 3 }{ n } -2-\cfrac { 1 }{ n } }{ 2+\cfrac { 1 }{ n } } ) } \\ =\cfrac { 3-2 }{ 2 } =\cfrac { 1 }{ 2 } $$
Question 3
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow 0 } \dfrac{1+\sin{x}-\cos{x}+\log{(1-x)}}{x^{3}}$$, is

A
$$-1$$

B
$$1/2$$

C
$$-1/2$$

D
$$1$$

Solution
Question 4
1 / -0

If $$ \mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) } $$ is non-
zero finite $$ ( \mathrm { a } > 0 ) , $$ then-

A
L = 2 , b = 1 , c = 1

B
$$

L = \frac { 1 } { 2 } , b = 1 , c = 1

$$

C
L = 4 , b = - 1 , c = - 1

D
$$

L = \frac { 1 } { 4 } , b = - 1 , c = - 1

$$
Question 5
1 / -0

The value of $$\displaystyle \lim_{\theta \rightarrow 0^{+}} \dfrac {\sin \sqrt {\theta}}{\sqrt {\sin \theta}}$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$4$$

Solution
Question 6
1 / -0

The value of $$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sqrt {1-\cos 2x}}{x}$$ equals

A
$$0$$

B
$$1$$

C
$$\sqrt {2}$$

D
$$Does\ not\ exist$$

Solution
Question 7
1 / -0

Let $$f:(0, \infty)\to R$$ be a differentiable function such that $$f'(x)=2-\dfrac{f(x)}{x}$$ for all $$x\in (0, \infty)$$ and $$f(1)\neq 1$$. Then

A
$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } f'\left( \dfrac { 1 }{ x } \right) =1$$

B
$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } xf\left( \dfrac { 1 }{ x } \right) =2$$

C
$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } x^{ 2 }f'\left( x \right) =0$$

D
$$\left| f\left( x \right) \right| \le 2$$ for $$ $$ all $$X\in \left( 0,2 \right) $$

Solution

Here, $$f'(x)=2-\dfrac{f(x)}{x}$$
or $$\dfrac{dy}{dx}+\dfrac{y}{x}=2$$ [ i.e. linear differential equation in $$y$$]
Integrating Factor, $$IF=e^{\displaystyle \int{\dfrac{1}{x}dx}}=e^{\log x}=x$$
$$\therefore$$ Required solution is $$y.(IF)=\displaystyle \int{Q(IF)dx+C}$$
$$\Rightarrow y(x)=\int{2(x)dx+C}$$
$$\Rightarrow yx=x^2+C$$
$$\therefore y=x+\dfrac{C}{x}$$ $$[ \because C \neq 0$$, as $$f(1) \neq 1]$$
(a) $$\displaystyle \lim_{x \rightarrow 0^+}{f' \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1-Cx^2)}=1$$
$$\therefore $$ Option (a) is correct.
(b) $$\displaystyle \lim_{x \rightarrow 0^+}{xf \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1+Cx^2)}=1$$
$$\therefore $$ Option (b) is correct.
(c) $$\displaystyle \lim_{x \rightarrow 0^+}{x^2f' (x)}=\displaystyle \lim_{x \rightarrow 0^+}{(x^2-C)}=-C\neq 0$$
$$\therefore $$ Option (c) is correct.
(d) $$f(x)=x+\dfrac{C}{x}, C \neq 0$$
For $$C > 0, \displaystyle \lim_{x\rightarrow 0^+}{f(x)=\infty}$$
$$\therefore $$ Function is not bounded in $$(0, 2)$$.
$$\therefore $$ Option (d) is incorrect.
Question 8
1 / -0

$$\displaystyle\lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\cot x-\cos x}{(\pi -2x)^3}$$ equals?

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

B
$$\dfrac{1}{24}$$

C
$$\dfrac{1}{16}$$

D
$$\dfrac{1}{8}$$

Solution
Question 9
1 / -0

For $$x>y$$, $$\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]}$$ is :

A
0

B
-1

C
1

D
2
Question 10
1 / -0

If the function $$f(x)$$ satisfies the relation $$f(x+y)=y\dfrac{|x-1|}{(x-1)}f(x)+f(y)$$ with $$f(1)=2$$, then $$\displaystyle\lim_{x\rightarrow 1}f'(x)$$ is?

A
$$2$$

B
$$-2$$

C
$$0$$

D
Limit do not exixst

Solution