MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 1

Evaluate

$\underset{x \rightarrow 3}\lim \sqrt[4] {x^3}$ using the properties of limits.

0%
$28^{1/4}$
0%
$25^{1/4}$
0%
$27^{1/4}$
0%
$26^{1/4}$

Explanation

$\displaystyle \lim _{ x\rightarrow 3 }{ \sqrt [ 4 ]{ { x }^{ 3 } } } \\={ (3^3) }^{ { 1 }/{ 4 } }={ (27) }^{ { 1 }/{ 4 } }$

$If {A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}}, \,i = 1,2,3,.....n$ and

${a_1}< {a_2}< {a_3}....< {a_{n,}} \, then$

$\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}......{A_n}} \right), 1 \le m \le n$

0%
is equal to ${\left( { - 1} \right)^m}$
0%
is equal to ${\left( { - 1} \right)^{m + 1}}$
0%
is equal to ${\left( { - 1} \right)^{n-m}}$
0%
is equal to ${\left( { - 1} \right)^{n-m-1}}$

$\displaystyle \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\frac{\pi }{2} - x} \right)}^3}}}$

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

$f(x)= x\sin\dfrac{1}{x} , \ for x\neq 0$

$= 0,\ for x=0$

Then.

0%
$f'(0^+) exit\ but \ f'(0^-)$ does not exit
0%
$f'(0^+) \ and \ f'(0^-)$ do not exit
0%
$f'(0^+) = f'(0^-)$
0%
none of these

$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } }$ is equal to

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

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right)\, = 4$ ,then

0%
$a=1,b=4$
0%
$a=1,b=-4$
0%
$a=2,b=-3$
0%
$a=2,b=3$

$\displaystyle\lim_{n\rightarrow \infty}\left(\tan\theta +\dfrac{1}{2}\tan \dfrac{\theta}{2}+\dfrac{1}{2^2}\tan \dfrac{\theta}{2^2}+...+\dfrac{1}{2^n}\tan\dfrac{\theta}{2^n}\right)$ equals?

0%
$\dfrac{1}{\theta}$
0%
$\dfrac{1}{\theta}-2\cot 2\theta$
0%
$2\cot 2\theta$
0%
None of these

Let p=

$\lim_{x\rightarrow 0+}(1+tan^{2}\sqrt{x})^{\frac{1}{2x}}$ then log p is equal to :

0%
1
0%
$\frac{1}{2}$
0%
$\frac{1}{4}$
0%
2

$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 4 } }{ { \left( \sin { 2x } \right) }^{ \sec ^{ 2 }{ 2x } } }$ is equal to

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

$\displaystyle \lim _{ \theta \rightarrow \pi /2 }{ \dfrac { 1-\sin \theta }{ (\pi /2-\theta )\cos { \theta } } }$ is equal to

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

$\underset {x \rightarrow \infty}{lim} (1+\frac {a}{x}-\frac {4}{x^{2}})^{2x} =e^{3}$ , then 'a' is equal to :

0%
2
0%
$\frac {3}{2}$
0%
$\frac {2}{3}$
0%
$\frac {1}{2}$

The value of

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

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

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

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

The value of

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

0%
$0$
0%
$1$
0%
$2$
0%
$\infty$

The value of

$\lim _{ x\rightarrow 0 }{ \dfrac { 1-\cos { ^{ 3 }x } }{ x\sin { x\cos { x } } } }$ is

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

The value of

$\displaystyle\lim _ { x \rightarrow \infty } a ^ { x } \sin \left( \frac { b } { a ^ { x } } \right)$ where

$a > 1$ is

0%
$b \log a$
0%
$a \log b$
0%
$b$
0%
$a$

The value of

$\displaystyle \lim _{ x\rightarrow 3 }{ \dfrac { \left( { x }^{ 3 }+27 \right) \log _{ e }{ \left( x-2 \right) } }{ { x }^{ 2 }-9 } }$ is

0%
$9$
0%
$18$
0%
$27$
0%
$\dfrac {1}{3}$

The value of

$\displaystyle \lim _{ x\rightarrow \frac { x }{ 4 } }{ \dfrac { 4\sqrt { 2 } -{ \left( \cos { x } +\sin { x } \right) }^{ 5 } }{ 1-\sin { 2x } } }$ is

0%
$0$
0%
$\sqrt {2}$
0%
$5\sqrt {2}$
0%
$3$

$\displaystyle \lim _ { x \rightarrow 0 } \left( \left[ \dfrac { - 5 \sin x } { x } \right] + \left[ \dfrac { 6 \sin x } { x } \right] \right)$ (where

$[ .]$ denotes greatest integer function) is equal to

0%
$0$
0%
$-12$
0%
$1$
0%
$2$

$\underset { x\rightarrow 0 }{ lim } \dfrac { { x }^{ 3 } }{ \sqrt { a+x } (bx-sinx) } =1,$ a > 0, then a + b is equal to

0%
36
0%
37
0%
38
0%
40

$\lim _ { x \rightarrow 0 } \dfrac { \sin x ^ { 5 } } { \sin ^ { 4 } x } =$

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

$\displaystyle\lim_{x\rightarrow 0}\dfrac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}=?$

0%
$10/3$
0%
$3/10$
0%
$6/5$
0%
$56$

$\underset { x\rightarrow 0 }{ lim } \dfrac { 1-\cos { 2x } }{ \cos { 2x } -\cos { 8x } }$ is equal to

0%
$-1/15$
0%
$1/10$
0%
$1/15$
0%
$15$

$\lim _ { x \rightarrow - 1 } \dfrac { \cos 2 - \cos 2 x } { x ^ { 2 } - | x | }$ is equal to :

0%
$0$
0%
$\cos 2$
0%
$2 \sin 2$
0%
None

$\lim _{ x\rightarrow 0 }{ \cfrac { x.{ 10 }^{ x }-x }{ 1-cosx } = }$

0%
$\log { 10 }$
0%
$2\log { 10 }$
0%
$3\log { 10 }$
0%
$4\log { 10 }$

$\lim _{ x\rightarrow 0 }{ \frac { \sqrt [ 3 ]{ 1+\sin { x } } -\sqrt [ 3 ]{ 1-\sin { x } } }{ x } } =$

0%
$0$
0%
$1$
0%
$\frac { 2 }{ 3 }$
0%
$\frac { 3 }{ 2 }$

$\underset { x\rightarrow 0 }{ lim } \frac { sin({ 6x }^{ 2 }) }{ Incos({ 2x }^{ 2 }-x) } =$

0%
12
0%
-12
0%
6
0%
-6

$\lim _ { x \rightarrow 1 } \left( \log _ { 3 } 3 x \right) ^ { \log _ { x } 3 } =?$

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

Let

$f ( \beta ) = \lim _ { \alpha \rightarrow \beta } \dfrac { \sin ^ { 2 } \alpha - \sin ^ { 2 } \beta } { \alpha ^ { 2 } - \beta ^ { 2 } },$ then

$f \left( \dfrac { \pi } { 4 } \right)$ is greater than-

0%
$\lim _ { x \rightarrow 0 } \dfrac { 1 - \cos ^ { 3 } x } { x \sin 2 x }$
0%
$\lim _ { x \rightarrow \pi / 2 } \dfrac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$
0%
$\lim _ { x \rightarrow \infty } ( \cos \sqrt { x + 1 } - \cos \sqrt { x } )$
0%
$\lim _ { x \rightarrow a } \dfrac { \sqrt { a + 2 x } - \sqrt { 3 x } } { \sqrt { 3 a + x } - 2 \sqrt { x } }$ where $a > 0$

$\underset { x\rightarrow -\infty }{ lim } \frac { ({ 3x }^{ 4 }+{ 2x }^{ 2 })sin(\frac { 1 }{ x } )+{ |x| }^{ 3 }+5 }{ { |x }|^{ 3 }+{ |x| }^{ 2 }+|x|+1 } =$

0%
2
0%
1
0%
-2
0%
-3

The value f

$\lim_{x\rightarrow \pi/4}\dfrac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}=$

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

$L = \underset{x \rightarrow 0}{lim} \dfrac{a \, sin \, x - sin \, 2x}{tan^3 x}$ is finite, then the value of L is :

0%
1
0%
2
0%
3
0%
-1

$\displaystyle \lim_{x\rightarrow 0}\dfrac {ae^{-x}-b\cos x-\dfrac {1}{2}cx}{x\cos x}=2$ then the value of

$a+b+c$ is-

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

$\displaystyle \lim_{x\rightarrow 0} \dfrac {\sin 2x + 3x}{2x + \sin 3x}$ is equal to

0%
$1$
0%
$\dfrac {1}{5}$
0%
$2$
0%
Does not exist

$f(x)$ is continuous and

$f\left(\dfrac {9}{2}\right)=\dfrac {2}{9}$ , then

$\displaystyle\lim _{ x\rightarrow 0 }f\left(\frac {1-\cos 3x}{x^2}\right)$ is equal to :

0%
$\dfrac {9}{2}$
0%
$\dfrac {2}{9}$
0%
0
0%
$\dfrac {8}{9}$

Explanation

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { 1-cos3x }{ x^{ 2 } } }$

$=\lim _{ x\rightarrow 0 }{ \dfrac { 2\sin ^{ 2 }{ (\frac { 3x }{ 2 } ) } }{ (\frac { 3x }{ 2 } )^{ 2 } } } \times \dfrac { 9 }{ 4 }$

$\displaystyle =\frac{9}{2}$

$\Rightarrow \displaystyle \lim _{ x\rightarrow 0 }{ f\left( \frac { 1-cos3x }{ x^{ 2 } } \right) } = f\left( \dfrac 9 2\right)=\dfrac { 2}{ 9 }$
.

The function

$f :( R-{0})$

$\rightarrow$ R given by

$\displaystyle f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at

$x = 0$ by defining

$f(0)$ as

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

Explanation

Given

$\displaystyle f\left( x \right) =\frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 }$

$\displaystyle \Rightarrow f\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \left\{ \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } \right\} } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 2x }-1-2x }{ x\left( { e }^{ 2x }-1 \right) } } \ ....... \quad \left[ \frac { 0 }{ 0 } form \right]$

$\therefore$ usingL'Hospital rule

$f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) } }$

$\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } } } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right) } =1$

If the function

$f(x)= \left\{\begin{matrix}\dfrac {\sqrt {2+\cos x}-1}{(\pi-x)^2} & x\neq \pi \\ k & x=\pi \end{matrix}\right.$ is continuous at

$x=\pi$ , then

$k$ equals :

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

Explanation

$\displaystyle\lim _{ x\rightarrow \pi }{ \frac { \sqrt { 2+\cos { x } } -1 }{ (\pi -x)^{ 2 } } } =\lim _{ x\rightarrow \pi }{ \frac { 1+\cos { x } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } }$ [Rationalize numerator and denominator ]

$\displaystyle =\lim _{ x\rightarrow \pi }{ \frac { 1-\cos { (\pi -x } ) }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } }$

$=\displaystyle \lim _{ x\rightarrow \pi }{ \frac { 2\sin ^{ 2 }{ \frac { (\pi -x) }{ 2 } } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } } =\frac { 1 }{ 4 }$

So,

$k =\dfrac{1}{4}$ .

$f:R\rightarrow R$ is a function defined by

$f(x)=[x]\displaystyle \cos\left(\frac{2x-1}{2}\pi\right)$ , where

$[{x}]$ denotes the greatest integer function, then

${f}$ is:

0%
continuous for every real $x$
0%
discontinuous only at $x = 0$
0%
discontinuous only at non-zero integral values of $x$
0%
continuous only at $x = 0$

Explanation

For

$n\in I$

$\displaystyle\lim _{ x\rightarrow n+ }{ f\left( x \right) } =\lim _{ x\rightarrow n+ }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =n\cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0$

And

$\displaystyle\lim _{ x\rightarrow n- }{ f\left( x \right) } =\lim _{ x\rightarrow n- }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =\left( n-1 \right) \cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0$

Thus

$f$ is continuous for

$x=n\in I$ ........ (1)

Since the function

$g\left( x \right)=\left[ x \right]$ and

$\displaystyle h\left( x \right) =\cos { \left( \frac { 2x-1 }{ 2 } \right) \pi }$ are continuous on

$x\in R-I$ , ......... (2)

From, (1) and (2), we get

$f$ is continuous everywhere

$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$

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

Explanation

$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at}$

$x\rightarrow 0$

$but\ have\ finite\ value = 0$

$0\times\ finite\ value\ =0$

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }$ is equal to

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

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }$

$=\displaystyle \lim_{x\to 0}\dfrac{xe^x}{x}-\lim_{x\to 0}\dfrac{\sin x}{x}$ , split up

$=\displaystyle \lim_{x\to 0}e^x-\lim_{x\to 0}\dfrac{\sin x}{x}$

$=e^0-1=1-1=0$

Use limit properties to evaluate

$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x$

0%
$12$
0%
$14$
0%
$16$
0%
$18$

Explanation

$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x =\lim_{x\to 4}\left(3x\tan\dfrac{\pi}{x}\right)$ , simplify

$=3(4)\tan\dfrac{\pi}{4}=12(1)=12$ , [substitute the limit ]

The from of the limit is not indeterminate, so we can substitute the limit value

For every integer

$n$ , let

$a_{n}$ and

$b_{n}$ be real numbers. Let function

$f: IR \rightarrow IR$ be given by

$f(x)=\left\{\begin{array}{l} a_{n}+\sin\pi x,\ for\ x\in[2n,\ 2n+1]\\ b_{n}+\cos\pi x,\ for\ x\ \in(2n-1,2n) \end{array}\right.$

$, for\ all\ integers\ n.$
lf

$f$ is continuous, then which of the following hold(s) for all

$n$ ?

0%
$a_{n-1}-b_{n-1}=0$
0%
$a_{n}-b_{n}=1$
0%
$a_{n}-b_{n+1}=1$
0%
$a_{n-1}-b_{n}=-1$

Explanation

Check the continuity at

$x=2n$

$L.H.L =\underset {x\rightarrow 2n^-}{\lim} [f(x)]$

$=\underset {h\rightarrow 0}{\lim}[f(2n-h)]$

$=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi (2n-h)]$

$=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi h]$

$=b_n+1$

$R.H.L=\underset {x\rightarrow 2n^+}{\lim}[f(x)]=\underset {h\rightarrow 0}{\lim}[f(2n+h)]$

$=\underset {h\rightarrow 0}{\lim}[a_n+\sin \pi(2n+h)]=a_n$

$f(2n)=a_n+\sin 2\pi n=a_n$

For continuity,

$\underset {x\rightarrow 2n^-}{\lim}f(x)=\underset {x\rightarrow 2n^+}{\lim} f(x)=f(2n)$

So,

$a_n=b_n+1\Rightarrow a_n-b_n=1$

Now, check for continuity at

$x=2n+1$

$L.H.L.=\underset {h\rightarrow 0}{\lim}(a_n+\sin \pi (2n+1-h))=a_n$

$R.H.L.=\underset {h\rightarrow 0}{\lim}(b_{n+1}+\cos (\pi (2n+1-h)))=b_{n+1}-1$

$f(2n+1)=a_n$ .

For continuity,

$a_n=b_{n+1}-1\Rightarrow a_{n-1}=b_n-1$

Both, option B and option D are correct.

$\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =A$ and

$\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =B$ , then which one of the following is correct?

0%
$A=1$ and $B=0$
0%
$A=0$ and $B=1$
0%
$A=0$ and $B=0$
0%
$A=1$ and $B=1$

Explanation

As given

$A=\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =\lim _{ x\rightarrow \infty }{ \cfrac { \sin { \left( \cfrac { 1 }{ x } \right) } }{ \left( \cfrac { 1 }{ x } \right) } }$
Let

$t=\cfrac { 1 }{ x } ;x\rightarrow \alpha ,t\rightarrow 0$

$\Rightarrow A=\lim _{ t\rightarrow \infty }{ \cfrac { \sin { t } }{ t } } =1\left[ \because \lim _{ t\rightarrow 0 }{ \cfrac { \sin { x } }{ x } =1 } \right]$
and

$B=\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } \Rightarrow B=0\quad$
Therefore,

$A=1$ and

$B=0$ is correct

The function

$f\left( x \right)=\left[ x \right] ,$ at

${ x }=5$ is:

0%
left continuous
0%
right continuous
0%
continuous
0%
cannot be determined

Explanation

We know that the greatest integer function is discontinuous at integer values. For a value of

$x$ just smaller than 5, the greatest integer smaller than it is 4. So,

$\lim _{ x\rightarrow { 5 }^{ - } }{ [x] } = 4$

$f\left( x=5 \right)=\left[ 5 \right] =5$
For a value of

$x$ just bigger than 5, the greatest integer smaller than it is 5. So,

$\lim _{ x\rightarrow { 5 }^{ + } }{ [x] } = 5$
Since

$\lim _{ x\rightarrow { 5 }^{ - } }{ f\left( x \right) } \neq f\left( 5 \right)$ , the function is not left continuous.
Since

$\lim _{ x\rightarrow { 5 }^{ + } }{ f\left( x \right) } =f\left( 5 \right)$ , the function is right continuous.
Correct answer: Option B

$f(x)=\left\{\begin{matrix} 2x-1& if &x>2 \\ k & if &x=2 \\ x^{2}-1& if & x<2\end{matrix}\right.$ is continuous at

$x= 2$ then

$k =$

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

Explanation

Since, the function is continuous at x = 2,

$\displaystyle\lim_{ x \rightarrow {2}^{-}} f(x) = f(2) = \displaystyle\lim _{x \rightarrow {2}^{+} }f(x)$

$\displaystyle\lim _{x \rightarrow {2}^{-}} 2x - 1 = f(2) = \displaystyle\lim_{ x \rightarrow {2}^{+}} {x}^{2} - 1$
k = 3
Hence, k = 3

Evaluate

$\displaystyle \lim_{x \rightarrow -2} \displaystyle \frac{x^2 - 1}{2x + 4}$ .

0%
0
0%
1
0%
2
0%
$\infty$

Explanation

Consider,

$\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}$

Substitute

$x=-2+h$

$\implies h\rightarrow 0 \text{ as } x\rightarrow -2$

$\implies \displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}$

$\displaystyle =\lim _{ h\rightarrow 0 } \frac { (h-2)^{ 2 }-1 }{ 2(h-2)+4 }$

$\displaystyle =\lim _{ h\rightarrow 0 } \frac { h^{ 2 }-4h+3 }{ 2h }$

$=\infty$

Hence,

$\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}=\infty$

$f(x)=\displaystyle \frac{\sin x}{x},x\neq 0$ is to be continuous at

${x}=0$ then

$\mathrm{f}({0})=$

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

Explanation

From the property of limits,

$\underset{ x \rightarrow 0}{Lim} \dfrac{sin x}{x} = 1$
Thus, for the function to be continuous at

$x = 0$

$f(0) = 1$

Say true or false.

Every continuous function is always differentiable.

0%
True
0%
False

Explanation

If a function

$f(x)$ is differentiable at a point

$a$ , then it is continuous at the point

$a$ . But the converse is not true.

$f(x)=|x|$ is contineous but not differentiable at

$x=0.$
Therefore, the given statement is false.

$\displaystyle \lim_{x\rightarrow \infty} \sin x$ equals

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

Explanation

Let

$L = \displaystyle \lim_{x\rightarrow \infty} \sin x$
Assume

$\displaystyle y = \frac{1}{x}$ so as

$x\to \infty, y \to 0$

$\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}$
We know

$\sin x$ lie between -1 to 1
so let

$p =\sin x$ as

$x\to \infty$
Thus left hand limit

$=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p$
and right hand limit

$=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p$
Clearly L.H.L

$\neq$ R.H.L

$\Rightarrow$ Required limit does not exist.

The function

$\displaystyle \mathrm{f}({x})=\frac{|x-3|}{x-3}$ at

${x}=3$ , is

0%
Left continuous
0%
Right continuous
0%
continuous
0%
discontinuous

Explanation

we just concentrate on the function

$\dfrac{|x|}{x}$
LHL = lim

$x \rightarrow {0}^{-} \dfrac{|x|}{x} = \dfrac{-x}{x} = -1$

RHL = lim

$x \rightarrow {0}^{+} \dfrac{|x|}{x} = \dfrac{x}{x} = 1$

Let ,

$y = x - 3$
Now, if

$x \rightarrow 3, then\ , y \rightarrow 0$
Applying the above derived result for y,
Since, the LHL is not equal to the RHL,
Hence the function is discontinuous at y = 0 or x = 3

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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