MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 9

$\underset { n \rightarrow \infty }{ Lt } \sum _{ r=2n+1\quad }^{ 3n } \dfrac {n}{r^2 - n^2}$ is equal to :

0%
$l n \sqrt{\frac {2}{3}}$
0%
$l n \sqrt{\frac {3}{2}}$
0%
$l n \frac {2}{3}$
0%
$l n \frac {3}{2}$

The value of

$\displaystyle \lim_{x \rightarrow 1^{-}}\dfrac {1 - \sqrt {x}}{(\cos^{-1}x)^{2}}$

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

Explanation

${ lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) \\ { \Rightarrow lim }_{ x\rightarrow 1 }1-\sqrt { x } =0\quad ,\quad { \Rightarrow lim }_{ x\rightarrow 1 }(cos^{ -1 }x)^{ 2 }=0\\ { lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) =\frac { 0 }{ 0 } \\ Now\quad using\quad l'hospital's\quad rule\\ { lim }_{ x\rightarrow 1 }\left( \dfrac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) ={ lim }_{ x\rightarrow 1 }\left( \dfrac { \frac { d(1-\sqrt { x } ) }{ dx } }{ \frac { d(cos^{ -1 }x)^{ 2 } }{ dx } } \right) ={ lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } \\ Now\quad { lim }_{ x\rightarrow 1 }1-\dfrac { 1 }{ 2\sqrt { x } } =1-1/2=1/2\\ And\quad { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } =\frac { 0 }{ 0 } \\ Again\quad using\quad l'hospital's\quad rule\\ \\ { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2d(cos^{ -1 }x)/dx }{ d(\sqrt { 1-{ x }^{ 2 } } )/dx } ={ lim }_{ x\rightarrow 1 }\dfrac { -2(\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } )\quad }{ \frac { -2x }{ 2\sqrt { 1-{ x }^{ 2 } } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2 }{ x } =-2\\ Therefore,\quad { lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } =\dfrac { 1/2 }{ -2 } =-1/4\\$

$\displaystyle \lim_{I\rightarrow \left (\dfrac {\pi}{2}\right )} = \int_{0}^{t}\tan \theta \sqrt {\cos \theta} ln (\cos \theta) d\theta$ is equal to

0%
$-4$
0%
$4$
0%
$-2$
0%
Does not exists

$f '$ (0) = 0 and f(x) is a differentiable and increasing function,then lim

$x \rightarrow 0$

$\frac {x.f ' (x^2)}{f ' (x)}$

0%
is always equal to zero
0%
may not exist as left hand limit may not exist
0%
may not exist as left hand limit may not exist
0%
right hand limit is always zero

Consider

$f(x)=\lim _{ n\rightarrow \infty }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } } }{ { x }^{ n }+\sin { { x }^{ n } } } }$ for

$x>0,x\neq 1,f(1)=0$ then

0%
$f$ is continuous at $x=1$
0%
$f$ has a discontinuity at $x=1$
0%
$f$ has an infinite or oscillatory discontinuity at $x=1$
0%
$f$ has a removal type of discontinuity at $x=1$

$\sum _{ r=1 }^{ k }{ \cos ^{ -1 }{ \beta } } =\cfrac { k\pi }{ 2 }$ for any

$k\ge 1$ and

$A=\sum _{ r=1 }^{ k }{ { \left( { \beta }_{ r } \right) }^{ r } }$ , then

$\lim _{ x\leftarrow A }{ \cfrac { { \left( 1+x \right) }^{ 1/3 }-{ \left( 1-2x \right) }^{ 1/4 } }{ x+{ x }^{ 2 } } }$ is equal to

0%
$0$
0%
$\cfrac{1}{2}$
0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { 5 }{ 6 }$

$\displaystyle \lim_{x \rightarrow 0} \frac{ae^x + b cos x + c. e^{-x}}{sin^2 x} = 4$ then b =

0%
2
0%
4
0%
-2
0%
-4

$\lim_{x\rightarrow \dfrac{\pi}{6}} \dfrac{2 sin^{2} x + sin x - 1}{2 sin^{2} x - 3 sin x + 1}$

0%
6
0%
-6
0%
-3
0%
3

Let f:R

$\rightarrow$ (0,1) be a continuous function.. Then, which of the following function(s) has (have) the value zero at some point in the interval (0,1)?

0%
$e ^ { x } - \int _ { 0 } ^ { 1 } f ( t ) \sin t d t$
0%
$f ( x ) + \int _ { 0 } ^ { 1 } f ( t ) \sin t d t$
0%
$x - \int _ { 0 } ^ { \frac { \pi } { 2 } - x } f ( t ) \cos t d t$
0%
$x ^ { 3 } - f ( x )$

$\mathop {\lim }\limits_{x \to {a^ + }} {{\left\{ x \right\}\sin \left( {x - a} \right)} \over {{{\left( {x - a} \right)}^2}}}$

is equal to (where {.} denotes the fraction
part of x and $a \in N$

0%
0
0%
1
0%
does not exist
0%
none of thes

$f\left( x \right) = \left\{ \begin{array}{l}\frac{{1 - \left| x \right|}}{{1 + x}},{\rm{ }}x \ne - 1\\1,{\rm{ }}x = - 1{\rm{ }}\end{array} \right.$ then

$f\left( {\left[ {2x} \right]} \right),$ where

$\left[ {} \right]$ represents the greatest integer function , is

0%
discontinuous at $x = - 1$
0%
continuous at $x = 0$
0%
continuous at $x = \frac{1}{2}$
0%
continuous at $x = 1$

$f : R \rightarrow R$ is defined by

$f(x) = \left \{\begin{matrix} \dfrac{x + 2}{x^2 + 3x + 2} & if & x \in R - \{-1, -1\} \\ -1 & if & x = -2 \\ 0 & if &x = -1\end{matrix} \right.$ then

$f(x)$ continuous on the set

0%
$R$
0%
$R - \{-2\}$
0%
$R - \{-1, -2\}$
0%
$R-\{-1\}$

$l=\lim\limits_{n\to 3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$ and

$m=\lim\limits_{n\to -3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$ , then

0%
$l\ne m$
0%
$l=2m$
0%
$l=-m$
0%
$l=m$

Explanation

Given

$l=\underset{x\to3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore,

$l=\underset {x\to3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to 3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$

Given

$m=\underset{x\to-3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore,

$m=\underset {x\to-3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to -3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to-3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$

Hence,

$l=8,m=8$

Therefore,

$l= m$

Let

$f(\theta) = \dfrac{1}{tan^{9}\theta} {(1+tan\theta)^{10}+(2+tan\theta)^{10}+....+(20+tan\theta)^{10}}-20tan\theta$ . The left hand limit of

$f(\theta)$ as

$\theta \rightarrow \dfrac{\pi}{2}$ is:

0%
1900
0%
2000
0%
2100
0%
2200

Consider

$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{bmatrix}$ , then the value of

$\lim_{n \rightarrow \infty} \dfrac{A^{n}}{n}$ (where

$\theta \in R$ ) is equal to

0%
$10$
0%
zero matrix
0%
symmetric matrix
0%
$4$

$\displaystyle\lim_ { x\rightarrow \lambda } { \left( 2-\dfrac { \lambda }{ x } \right) }^{ \lambda tan\left( \dfrac { \pi x }{ 2\lambda } \right) }=\frac { 1 }{ e } ,$ then

$\lambda$ is equal to-

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

$\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right)$ has the value equal to

0%
$\dfrac {a-b}{3}$
0%
$0$
0%
$\dfrac {(a^{2}-b^{2})}{6a^{2}b^{2}}$
0%
$\dfrac {a^{2}-b^{2}}{3a^{2}b^{2}}$

$\displaystyle \lim _{ x\rightarrow 2 }{ \frac { \sqrt [ 3 ]{ 60+{ x }^{ 2 } } -4 }{ \sin { \left( x-2 \right) } } }$

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

$if\left( x \right)$ = greatest integer

$\le x$ , then

$\underset { x\longrightarrow 2 }{ lim } { \left( -1 \right) }^{ \left[ x \right] } is equal to-$ is equal to-

0%
0
0%
-1
0%
+1
0%
none of these

The value of

$\displaystyle \lim _{ x\rightarrow 0 }{ \csc^{ 4 }{ x } \int _{ 0 }^{ { x }^{ 2 } }{ \frac { ln\left( 1+4t \right) }{ { t }^{ 2 }+1 } } dt }$ is

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

The value of

$\displaystyle\lim_{n\rightarrow \infty}n(n\{ln (n)-ln (n+1)\}+1)$ is?

0%
e
0%
$\dfrac{1}{e}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

$\displaystyle \lim_{x\rightarrow \infty}{x^{2}\sin\left(\log_{e}\sqrt{\cos\dfrac{\pi}{x}}\right)}$

0%
$0$
0%
$-\dfrac{\pi^{2}}{2}$
0%
$-\dfrac{\pi^{2}}{4}$
0%
$-\dfrac{\pi^{2}}{8}$

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

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

Explanation

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,

$=\dfrac{1}{16}$

$\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } \frac { (1-sinx)({ 8x }^{ 2 }-{ \pi }^{ 3 })cosx }{ { (\pi -2x) }^{ 4 } }$

0%
$-\cfrac { { \pi }^{ 2 } }{ 16 }$
0%
$\cfrac { { 3\pi }^{ 2 } }{ 16 }$
0%
$\cfrac { { \pi }^{ 2 } }{ 16 } \quad$
0%
$-\cfrac { 3{ \pi }^{ 2 } }{ 16 } \quad$

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

0%
$\cfrac { \pi} {180}$
0%
$\cfrac { \pi} {90}$
0%
${0}$
0%
$\cfrac {90} { \pi }$

The value of

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

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

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

$a,b,c$ are respectively

0%
$0,6,6$
0%
$6,0,6$
0%
$6,6,0$
0%
$0,0,6$

Explanation

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.

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

0%
$\sqrt [ 3 ]{ a\quad cos\quad a }$
0%
$2\sqrt [ 3 ]{ a }$
0%
${ 3a }^{ { 2 /}{ 3 } }cos\quad a$
0%
$\sqrt [ 2 ]{ a\quad cos\quad a }$

The value of

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

0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {3}{2}$
0%
$\dfrac {3}{4}$

Integrate:

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

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

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

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

The value of

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

0%
$0$
0%
$\infty$
0%
$e^{1/4}$
0%
$Does\ not\ exist$

The value of

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

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

$\lim_{n\rightarrow \infty}\dfrac{1}{n^{2}}\left[\sin^{3}\dfrac{\pi}{4n}+2\sin^{3}\dfrac{2\pi}{4n}+3\sin^{3}\dfrac{3\pi}{4n}+....+n\sin^{3}\dfrac{n\pi}{4n}\right]=$

0%
$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52-15\pi\right)$
0%
$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52+15\pi\right)$
0%
$\dfrac{\sqrt{2}}{9\pi}\left(52-17\pi\right)$
0%
$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52+17\pi\right)$

$\alpha \quad and \beta$ are the roots of the equation

${ax}^{2}+bx+c=0$ , then

$\underset { x\rightarrow \cfrac { \pi }{ 2 } }{ lim } \cfrac { tan\left[ \left( \alpha +\beta \right) x \right] }{ sin\left[ \left( \alpha \beta \right) x \right] }$ is equal to :

0%
$\cfrac {c} {b}$
0%
$- \cfrac {b} {c}$
0%
$\cfrac {a} {b}$
0%
$- \cfrac {a} {b}$

$\begin{matrix} lim \\ n\rightarrow \infty \end{matrix}\int _{ 0 }^{ 1 }{ \frac { { nx }^{ n-1 } }{ 1+{ x }^{ 2 } } } dx=$

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

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)

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

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

Explanation

(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)

$\underset { x\rightarrow 0 }{ lt } \dfrac { { log }_{ e }\left( 1+x \right) }{ x } .\underset { x\rightarrow 0 }{ lt } \dfrac { 1 }{ 1+x } =1$

$\therefore$

$3<4<1<2$ .

$\mathop{\lim}\limits_{x \to 0} \left(\dfrac{3+x}{3-x}\right)^{\dfrac{x+1}{x}}$ is equal to

0%
$e^{2/3}$
0%
$e^{1/3}$
0%
$e^3$
0%
$e^2$

$\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 1,$ then

0%
$a = \frac{5}{2}$
0%
$b = \frac{{ - 5}}{2}$
0%
$a + b = 4$
0%
$a + b = -4$

$\lim_{x\rightarrow 0 }(\frac{p^{\frac{1}{x}}+q^{\frac{1}{x}}+r^{\frac{1}{x}}+s^{\frac{1}{x}}}{4})3x$ where p,q,r,s

$> 0$ is equal to

0%
$pqrs$
0%
$(pqrs)^{3}$
0%
$(pqrs)\frac{3}{2}$
0%
$(pqrs)\frac{3}{4}$

$\lim- {x\to 0}$

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

0%
4
0%
16
0%
32
0%
None of these

The value of

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

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

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

$a$

0%
$\dfrac { 1 }{ n }$
0%
$n-\dfrac { 1 }{ n }$
0%
$n+\dfrac{1}{n}$
0%
$None\ of\ these$

$\lim _ { x \rightarrow 0 } \frac { 1 - \cos x \cos 2 x \cos 3 x } { \sin ^ { 2 } 2 x } =$

0%
$\frac { 3 } { 2 }$
0%
$\frac { 5 } { 2 }$
0%
$\frac { 7 } { 4 }$
0%
$\frac { 9 } { 2 }$

The value of

$lim_{x \to 0} (\dfrac{1}{x^2} - cotx)$ equals

0%
$1$
0%
$0$
0%
$\infty$
0%
Does not exist

$\displaystyle \lim _{ x-\infty }{ sgn\left( \cot{\dfrac { { \pi x }^{ 2019 } }{ { x }^{ 2019 }+7 }} \right) }$

0%
Equals $-1$
0%
Equals $1$
0%
equals $0$
0%
Does not exit

$\displaystyle\lim_{x \to \pi/2} (sec x +tan x)$ is equal to

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

$\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)=$

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

$\underset { x\rightarrow 0 }{ lim } \dfrac { x\tan { 2x } -2\tan { 2x } }{ { \left( 1-cos2x \right) } }$ equals:

0%
$\dfrac{1}{4}$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

$\displaystyle\lim_{x\to \pi/2} \dfrac{sinx-(sinx)^{sin x}}{1-sin x + In sin x}$ is equal to

0%
$4$
0%
$2$
0%
$1$
0%
none of these

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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