MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 10

$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$ is equal to

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Evaluate :

$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right) }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x } }{ { e }^{ x\ell nx } } \right) }$ is equal to

0%
$\dfrac{1}{e}\ell n3$
0%
$e\ \ell n\ 3$
0%
$3$
0%
$\dfrac{1}{3}$

$\underset{x \rightarrow 0}{Lt} \dfrac{sin x - x + x^3 / 6}{x^5}$ =

0%
$\dfrac{1}{120}$
0%
$\dfrac{1}{110}$
0%
$\dfrac{1}{100}$
0%
none of these

$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$ is equal to

0%
$\cos^{4}\alpha.\log(\cos\alpha)-\sin^{4}\alpha.\log(\sin\alpha)$
0%
$\sin^{4}\alpha.\log(\cos\alpha)-\cos^{4}\alpha.\log(\sin\alpha)$
0%
$\sin^{4}\alpha.\log(\cos\alpha)+\cos^{4}\alpha.\log(\sin\alpha)$
0%
$None\ of\ the \ above$

$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { 3\sin { \left( { x }^{ 9} \right) -\sin { \left( { x }^{ 9 } \right) } } }{ { x }^{ 3 } } } =q$

0%
0
0%
$4\left(\dfrac{\pi}{200}\right)^{3}$
0%
$\dfrac{\pi}{200}$
0%
$\dfrac{\pi}{100}$

$\underset { x\rightarrow 0 }{ Lt } \cfrac {tanx-x}{x^2tanx}$ equals:

0%
1
0%
1/2
0%
1/3
0%
None of these

evaluate

$\underset { x\rightarrow 0 }{ lim } \frac { x-\int _{ 0 }^{ x }{ { cost }^{ 2 }dt } }{ { x }^{ 3 }-6x }$

0%
$3$
0%
$-1$
0%
$0$
0%
$1$

$\lim _ { x \rightarrow 0 } \left\{ \tan \left( \frac { \pi } { 4 } + x \right) \right\} ^ { \frac { 1 } { x } } =$

0%
$e$
0%
$e ^ { 2 }$
0%
$e ^ { - 1 }$
0%
$e ^ { - 2 }$

$\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x+1}{2x+1}\right)^{x^2}$ equals?

0%
$0$
0%
e
0%
$1$
0%
$\infty$

$g(x)=\frac { x }{ \left[ x \right] } for\quad x>2\quad then\quad \underset { x\rightarrow { 2 }^{ + } }{ Lim } \frac { g\left( x \right) -g\left( 2 \right) }{ x-2 }$

0%
-1
0%
0
0%
1/2
0%
1

$\underset{x \rightarrow \infty}{lim} \dfrac{2 \tan^{-1} x}{\pi}$ equals

$e^L$ then

$L$ is equal to

0%
$\dfrac{2}{\pi}$
0%
$-\dfrac{2}{\pi}$
0%
$-\dfrac{\pi}{2}$
0%
$1$

$\displaystyle \lim _{ x\rightarrow x/2 } \dfrac { \left[ 1-\tan { \left( \dfrac { x }{ 2 } \right) } \right] \left[ 1-\sin { x } \right] }{ \left[ 1+\tan { \left( \dfrac { x }{ 2 } \right) } \right] \left[ \pi -2x \right] ^{ 3 } }$ is

0%
$\dfrac {1}{8}$
0%
$0$
0%
$\dfrac {1}{32}$
0%
$\infty$

$\displaystyle \lim _{ \theta \rightarrow 0 }{ \frac { 4\theta \left( \tan { \theta -2\theta \tan { \theta } } \right) }{ { \left( 1-\cos { 2\theta } \right) }^{ 2 } } }$ is

0%
$1/\sqrt {2}$
0%
$1/2$
0%
$1$
0%
$2$

the value of

$\underset { x\rightarrow \infty }{ lim } \frac { { x }^{ 5 } }{ { 5 }^{ x } }$ is-

0%
0
0%
1
0%
${ e }^{ 5 }$
0%
${ e }^{ -5 }$

$k$ is an integer such that

$\lim_{n \rightarrow \infty} \left[\left(\cos \dfrac{k\pi}{4}\right)^{2}-\left(\cos \dfrac{k\pi}{6}\right)^{2}\right]=0$ then :

0%
$k$ is divisible neither by $4$ nor by $8$ .
0%
$k$ must be divisible by $12$ but not necessarily by $24$ .
0%
$k$ must be divisible by $24$ .
0%
either $k$ is divisible by $24$ or $k$ is neither divisible by $4$ nor by $6$ .

$\lim _ { x \rightarrow 5 } \frac { \sin ^ { 2 } ( x - 5 ) \tan ( x - 5 ) } { \left( x ^ { 2 } - 25 \right) ( x - 5 ) } =$

0%
1 $/ 10$
0%
$1$
0%
$0$
0%
$-6$

The value of

$\underset{x \rightarrow 0}{lim} \cos \,ec^4 \,x \displaystyle \overset{x^2}{\underset{0}{\int}} \dfrac{In(1 + 4t)}{1 + t^2}dt$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

$\lim_{n \rightarrow \infty}n^{2}\left(x^{1/n}-x^{1/\left(n+n\right)}\right),x>0$ , is equal to

0%
$0$
0%
$e^{x}$
0%
$\log_{e}x$
0%
$none\ of\ these$

$\lim_{x\rightarrow 0}\frac{ln(sin 3x)}{ln(sin x)}$ is equal to

0%
0
0%
1
0%
2
0%
Non existent

$\underset{x \rightarrow n/2}{lim} \dfrac{\cot x - \cos x}{(\pi - 2x)^3}$ equals

0%
$\dfrac{1}{24}$
0%
$\dfrac{1}{16}$
0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{4}$

$\underset{x \rightarrow 2}{lim} \dfrac{\sqrt[3]{60 + x^2} - 4}{\sin (x - 2)}$ equals

0%
$\dfrac{1}{4}$
0%
$0$
0%
$\dfrac{1}{12}$
0%
Does not exist

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

0%
$\frac { 3 } { 4 }$
0%
$\frac { 1 } { 4 }$
0%
$\frac { 4 } { 3 }$
0%
$0$

$\lim_{x\rightarrow 0}\dfrac{2\left(\sqrt{3}\sin\left(\dfrac{\pi}{6}+x\right)-\cos\left(\dfrac{\pi}{6}+x\right)\right)}{x\sqrt{3}\left(\sqrt{3}\cos x-\sin x\right)}$

0%
$-1/3$
0%
$2/3$
0%
$4/3$
0%
$-4/3$

$\lim_{x\rightarrow \pi/4}\dfrac{2\sqrt{2}-\left(\cos x+\sin x\right)^{2}}{1-\sin 2x}$ is equal to

0%
1
0%
$2\sqrt{2}$
0%
$\dfrac{4\sqrt{2}}{3}$
0%
$does\ not\ exist$

Let

$f$ be a differentiable function such that

$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$ and

$f(1) \neq 4$ .
Then

$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right)$ :

0%
Exists and equals 4
0%
Does not exist
0%
Exist and equals
0%
Exists and equals $\dfrac{4}{7}$

Explanation

$f'(x) = 7 -\dfrac{3}{4}\dfrac{f(x)}{x}$

$(x > 0)$

Given

$f(1) \neq 4$

$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = ?$

$\dfrac{dy}{dx} + \dfrac{3}{4}\dfrac{y}{x} = 7$ (This is LDE)

$=e^{\int \tfrac{3}{4x}dx} = e^{\tfrac{3}{4}ln|x|} = x^{\tfrac{3}{4}}$

$y.x^{\frac{3}{4}} =\int 7.x^{\frac{3}{4}} dx$

$y.x^{\frac{3}{4}} = 7.\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$

$f(x) = 4x+C.x^{-\frac{3}{4}}$

$f\left(\dfrac{1}{x}\right) = \dfrac{4}{x} + C.x^{\frac{3}{4}}$

$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = \underset{x\to 0^+}{\lim}\left(4+C.x^{\frac{7}{4}}\right) = 4$

$\underset { x\rightarrow 0 }{ Lim } \dfrac { \left( 1+{ a }^{ 3 } \right) +8{ e }^{ 1/x } }{ 1+\left( 1-{ b }^{ 3 } \right) +{ e }^{ 1/x } } =2$ then

0%
$a=1,b=2$
0%
$a=1,b={-3}^{1/3}$
0%
$a=2,b={3}^{1/3}$
0%
none of these

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

0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{12}$
0%
$\dfrac{5}{12}$

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

0%
$1$
0%
$0$
0%
$-1$
0%
$\frac { 1 } { 2 }$

Evaluate

$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \sin { \left[ \cos { x } \right] } }{ 1+\left[ \cos { x } \right] } }$ (

$[.]$ denotes the greatest integer function)

0%
$-1$
0%
$1$
0%
$0$
0%
$does\ not\ exist$

$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ \sin { x } }-1 \right) }{ 1-\cos { x } } }$

0%
$\dfrac {1}{2}$
0%
$2$
0%
$0$
0%
$1$

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

0%
$P=1/4$
0%
$P=1/2$
0%
$P=2$
0%
$P=1$

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

0%
$n!$
0%
$1$
0%
$\dfrac{1}{n!}$
0%
$0$

Explanation

$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { { 1 }^{ 1/x }+{ 2 }^{ 1/x }+{ 3 }^{ 1/x }......+{ n }^{ 1/x } }{ n } \right) }^{ nx }$

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

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?

0%
$1$
0%
$0$
0%
$-1$
0%
$\infty$

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

0%
$\dfrac{1}{4}$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$None of these$

Let

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

0%
$\displaystyle\lim _{ x\rightarrow \infty }{ xf\left( x \right) }$
0%
$\displaystyle\lim _{ x\rightarrow 1 }{ f\left( x \right) }$
0%
$\displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right) }$
0%
$\displaystyle\lim _{ x\rightarrow -\infty }{ f\left( x \right) }$

$\lim _{x \rightarrow 0}\left(\cos x+a^{3} \sin \left(b^{6} x\right)\right)^{\frac{1}{x}}=e^{512}$ then value of

$ab^2$ is equal to

0%
$-512$
0%
$512$
0%
$8$
0%
none of these

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

0%
-2
0%
-1
0%
0
0%
2

$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { { \left( \cos { x } \right) }^{ 1/2 }-{ \left( \cos { x } \right) }^{ 1/3 } }{ \sin ^{ 2 }{ x } } }$ is $$

0%
$1/6$
0%
$-1/12$
0%
$2/3$
0%
$1/3$

The value of

$\begin{matrix} lim \\ x\rightarrow y \end{matrix}\dfrac { { sin }^{ 2 }x-sin^{ 2 }y }{ { x }^{ 2 }-{ y }^{ 2 } }$

0%
0
0%
1
0%
$\dfrac { siny }{ y }$
0%
$\dfrac { sin2y }{ 2y }$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$ is equal to?

0%
$2$
0%
$-2$
0%
$1/2$
0%
$-1/2$

$\underset { x\rightarrow 0 }{ lim } (\cos x+a\sin b{ x) }^{ \frac { 1 }{ x } }$ is equal to

0%
$e^a$
0%
$e^{ab}$
0%
$e^b$
0%
$e^{a/b}$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{x(1+a\cos x)-b\sin x}{x^{3}}}=1$ then

0%
$a=-5/2$
0%
$a=-3/2,b=-1/2$
0%
$a=-3/2,b=-5/2$
0%
$a=-5/2,b=-3/2$

$\displaystyle\lim_{x\rightarrow 0}[1+x ln (1+b^2)]^{1/x}=2b\sin^2\theta, b >$ s m f

$\theta\in (-\pi, \pi]$ , then the value of

$\theta$ is?

0%
$\pm \dfrac{\pi}{4}$
0%
$\pm \dfrac{\pi}{3}$
0%
$\pm \dfrac{\pi}{6}$
0%
$\pm \dfrac{\pi}{2}$

$\lim _ { x \rightarrow \infty } \dfrac { \sin \left( \pi \cos ^ { -2 } x \right) } { x ^ { 2 } }$ is equal to:

0%
$- \pi$
0%
$\pi$
0%
$\dfrac { \pi } { 2 }$
0%
$1$

The value of

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

0%
$0$
0%
$1$
0%
$-1$
0%
$4$

Let

$f$ :

$\left ( \dfrac{\pi}{2}, \dfrac{\pi}{2} \right )\rightarrow R$ ,

$f(x) = \left \{\begin{matrix} \lim_{n\rightarrow \infty }\dfrac{(tanx)^{2n} + x^2}{sin^2x + (tanx)^{2n}}; & x \neq 0 \\ 1; & x = 0 \end{matrix} \right., n \in N$ . Which of the following holds good ?

0%
$f \left( -\dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)$
0%
$f \left( -\dfrac{\pi^-}{4} \right) = f \left (-\dfrac{\pi^+}{4} \right)$
0%
$f \left( \dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)$
0%
$f(0^+) = f(0) = f(0^)$

$\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?$

0%
$2_{\displaystyle /3}$
0%
$0$
0%
$1_{\displaystyle /3}$
0%
$-2_{\displaystyle /3}$

Consider

$\lim _ { x \rightarrow 0 } \dfrac { a x + b e ^ { - x } + \sin x + 1 } { a x - b \sin x } = \ell ( \ell$ is a finite number)

0%
$\ell = 0 , \text { if } a = - 2 , b = - 1$
0%
$\ell = 5 , i f a = 1 , b = - 1$
0%
$\ell = 1 , \text { if } a = b = - 1$
0%
None of these

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin(\pi \cos^2x)}{x^2}$ is equal to?

0%
$-\pi$
0%
$\pi$
0%
$\pi/2$
0%
$1$

$f(x)$ is odd linear polynomial with

$f(1)=1$ , then

$\underset{x \to 0} {\lim} \dfrac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}$ is

0%
$1$
0%
$\ln 2$
0%
$\dfrac{1}{2}\ln 2$
0%
$\cos 2$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 10 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 10 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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