MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 2

$x$ is very large, then

$\dfrac {2x}{1+x}$ is

0%
close to $0$
0%
arbitrarily large
0%
lie between $2$ and $3$
0%
close to $2$

Explanation

$\dfrac {2x}{1+x}=\dfrac {2}{1+\dfrac {1}{x}}=\dfrac{2}{1+0}=2$

$x\rightarrow \alpha$ then

$\dfrac {1}{x}\rightarrow 0$ .

What is

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } }$ equal to?

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

Explanation

Here, putting

$x=0$ directly we have,

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } }$

$=\dfrac {\cos 0 }{ \pi -0 } =\dfrac { 1 }{ \pi }$

Hence, C is correct.

$\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{b}{x}}}$ is equal to

0%
$e^{ab}$
0%
$e^{a+b}$
0%
non-existent
0%
none of these

Explanation

Now,

$\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{b}{x}}}$

$=\underset{x \rightarrow 0}{lim}\left\{(1+ax)^{\large{\frac{1}{ax}}}\right\}^{ab}$

$=\left\{\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{1}{ax}}}\right\}^{ab}$

$=e^{ab}$ . [ Since

$\lim\limits_{x \to 0}(1+x)^{\large{\frac{1}{x}}}=e$ ].

$\displaystyle{\lim_{x \to 0}}$

$\Bigg(\dfrac{(1+x)^{2}}{e^{x}}\Bigg)^\dfrac{4}{\sin x}$ is:

0%
$e^2$
0%
$e^{4}$
0%
$e^8$
0%
$e^{-8}$

Explanation

Now

$\lim\limits_{x \to 0}\left(\dfrac{(1+x)^2}{e^{x}}\right)^{\dfrac{4}{\sin x}}$

$=\lim\limits_{x \to 0}\dfrac{\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\dfrac{x}{\sin x}}\right)^8}{e^{\cfrac{4x}{\sin x}}}$

We have

$\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$ and

$\lim\limits_{x \to 0}(1+x)^{\dfrac{1}{x}}=e$ .

As both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing, so we can write the above limits as

$=\dfrac{\lim\limits_{x\to 0}\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\lim\limits_{x\to 0}\dfrac{x}{\sin x}}\right)^8}{\lim\limits_{x\to 0}e^{\cfrac{4x}{\sin x}}}$ [Using division property of limits]

$=\dfrac{\left(\lim\limits_{x\to 0}\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\left(\lim\limits_{x\to 0}\dfrac{x}{\sin x}\right)}\right)^8}{e^{\left(\lim\limits_{x\to 0}\dfrac{4x}{\sin x}\right)}}$ [Using limit property]

$=\dfrac{(e^1)^8}{e^{\frac{4}{1}}}=e^2$

If a sequence

$< a_{n} >$ is such that

$a_{1},a_{n+1}=\dfrac {2+3a_{n}}{1+2a_{n}}$ and

$\displaystyle \lim_{n \rightarrow \infty}a_{n}$ exists, then

$a_{n}$ is equal to

0%
$\cos 36^{o}$
0%
$2\cos 36^{o}$
0%
$\sin 18^{o}$
0%
$2\sin 36^{o}$

$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$ equals

0%
$\dfrac{7}{3}$
0%
$\dfrac{10}{3}$
0%
$\dfrac{14}{3}$
0%
$\dfrac{1}{3}$

Explanation

We have,

$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$

$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7x}{\dfrac{\sin3x}{3x}\times 3x}$

$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7}{\dfrac{\sin3x}{3x}\times 3}$

$=\dfrac{1\times 7}{1\times 3}$

$=\dfrac{7}{3}$

Hence, this is the answer.

$\displaystyle \lim_{x\rightarrow 0}{(1+\sin x)^{\cos x}}$ is equal to

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

Explanation

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

$={\left(1+\sin{0}\right)}^{\cos{0}}$

$={1}^{1}=1$

Evaluate the following limit :

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

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

Explanation

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

$=lim_{x\rightarrow 0} \dfrac{2\sin^2 x}{x^2}$

$=2lim_{x \rightarrow 0} (\dfrac{\sin x}{x} \times \dfrac{\sin x}{x})$

$=2lim_{x\rightarrow 0} \dfrac{\sin x}{x} \times lim_{x\rightarrow 0} \dfrac{\sin x}{x}=2(1)(1)=2$

Evaluate

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$

0%
$\cfrac { 1 }{ 2 }$
0%
$1$
0%
${ 3 }^{ 2 }$
0%
$\cfrac { 1 }{ { 2 }^{ 2 } }$

Explanation

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$

$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}$

$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{2}(1+\dfrac{1}{n})=\dfrac{1}{2}$

Which of the following statement is not correct

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$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } }$

Explanation

All the result other than D are correct.

The last result will hold only when

$g(x)\ne 0$ and

$\lim\limits_{x\to c} g(x)\ne 0$ for

$x$ belonging to a neighbourhood of

$c$ .

A, B, C are general results which are always true.

What is the value of

$\underset{x\rightarrow 0}{lim}\dfrac{\sin\,x}{\tan \,3x}$

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$1$

Explanation

Given,

$\lim_{x\rightarrow 0}\dfrac{\sin \left(x\right)}{\tan \left(3x\right)}$

applying L'Hospital rule, we get,

$\lim_{x\rightarrow 0} \dfrac{\cos \left(x\right)}{\sec ^2\left(3x\right)\cdot 3}$

$=\dfrac{\cos \left(0\right)}{\sec ^2\left(3\cdot 0\right)\cdot 3}$

$=\dfrac{1}{3}$

$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$
Then

$f(2^{+})-f(2^{-})=$

0%
$0$
0%
$2$
0%
$9$
0%
$4$

Explanation

Given:

$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$

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

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

$f(2^+)=16$

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

$=\displaystyle \lim _{ h\rightarrow 0}4(2-h)^{2}-9$

$f(2^-)=7$

So,

$f(2^{+})-f(2^{-})=16-7=9$

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\\mathrm{x}+\mathrm{d}(\mathrm{x}\geq\alpha)\end{array}\right.$ is such that

$\displaystyle \lim_{x\rightarrow \alpha}f(x) =L$ , then

$L=$

0%
$2d-b$
0%
$b-d$
0%
$d+b$
0%
$b-2d$

Explanation

$\begin{matrix}lim\\h\rightarrow 0^{-} \end{matrix}\ f(x)=2(\alpha -h)+b=2\alpha +b=L .............(1)\$

$\begin{matrix}lim\\h\rightarrow 0 ^{+} \end{matrix}\ f(x)=(\alpha +h)+d=L\quad \quad \ \alpha =L-d\ ................(2)$

Substituting value of euation (2)in (1), we get

$2\left ( L-d \right )+b=L$

$L=2d-b$

Hence, option 'A' is correct.

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$

0%
$\pi$
0%
$2\pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{2}{\pi}$

Explanation

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$

$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) }$
It is of the form

$\displaystyle \dfrac{0}{0}$ , so applying L-Hospital's rule

$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } }$

$=\displaystyle \dfrac{2}{\pi}$

$f(x)=x,x<0;f(x)=0,x=0$ ;

$f(x)=x^{2};x>0$ , then

$\displaystyle \lim_{x\rightarrow 0}f(x)$ is equal to

0%
Does not exist
0%
$0$
0%
$-1$
0%
$1$

Explanation

Given:

$f(x)$ =

$\left\{ \begin{matrix} x \\ 0 \\ x^ 2 \end{matrix}\begin{matrix}\quad x<0 \\\quad x=0 \\ \quad x>0 \end{matrix} \right\}$

To Find :

$\underset{x\rightarrow 0}{\lim}$

$f(x)$ = ?

Sol: left hand limit

$\rightarrow$

$\underset{x\rightarrow 0^-}{\lim}$

$f(x)$ =

$\underset{x\rightarrow 0}{\lim}$

$x$ =

$0$

right hand limit

$\rightarrow$

$\underset{x\rightarrow 0^+}{\lim}$ f(x)=

$\underset{x\rightarrow 0}{\lim}$

$x^2$ =

$0$ ,

LHL=RHL

$f(0)=0$

$LHL=RHL=f(0)=0$

$f(x)=2x+1, a=1,l=3$ and

$\epsilon=0.001$ , then

$\delta>0$ satisfying

$0<|x-a|<\delta$ such that

$|f(x)-l|<\epsilon$ , is

0%
$0.0005$
0%
$0.005$
0%
$0.001$
0%
$0.0001$

Explanation

We have to find

$\delta>0$ , which satisfies

$0<|x-a|<\delta$ such that

$|f(x)-l|<\epsilon$

$\because |f(x)-l| <\epsilon$

$\Rightarrow -\epsilon <f(x)-l<\epsilon$

$\Rightarrow l-\epsilon<f(x)<l+\epsilon$

$\Rightarrow 3-\epsilon<2x+1<3+\epsilon$

$\Rightarrow -\epsilon<2x-2<\epsilon$

$\Rightarrow -\dfrac{\epsilon }{2}<x-1<\dfrac{\epsilon }{2}$

$\Rightarrow |x-1|<\dfrac{\epsilon }{2}$

$\Rightarrow |x-a| <\dfrac{\epsilon }{2}$

$\delta = \dfrac{\epsilon}{2} = \dfrac{0.001}{2} = 0.0005$

Hence, option A.

Evaluate:

$\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)$

0%
$\displaystyle \frac{1}{12}$
0%
$-\displaystyle \frac{4}{3}$
0%
$-\displaystyle \frac{16}{3}$
0%
$-\displaystyle \frac{1}{48}$

Explanation

Let

$L =\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right) =\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{\sqrt[3]{8+h}}-\frac{1}{2}\right)$

Let

$a= \sqrt[3]{8+h}\Rightarrow a^3 = 8+h$ and

$b=2\Rightarrow b^3 =8$

Also

$a^3-b^3=(a-b)(a^2+b^2+ab)=8+h-8 =h\Rightarrow$ (1)

$\therefore\displaystyle L = \lim_{h\to 0}\frac{1}{h}\left( \frac{1}{a}-\frac{1}{b}\right)=-\lim_{h\to 0}\frac{a-b}{h}\cdot \frac{1}{ab}$

$\displaystyle \quad = -\lim_{h\to 0}\frac{a^3-b^3}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$

$\displaystyle \quad = -\lim_{h\to 0}\frac{h}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$ .....using

$(1)$

${h \rightarrow 0}\Rightarrow a^3=8\Rightarrow a=2$

$\displaystyle \quad =-\frac{1}{2\times 2(2^2+2^2+2\times 2)}=$

$=-\dfrac{1}{48}$

$\displaystyle \lim_{x\rightarrow a^+}f(x)=L$ , then for each

$\epsilon>0$ , there exists

$\delta>0$ so that

0%
$0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon$
0%
$0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$
0%
$a < x < a+\delta\Rightarrow f(x) - L <\epsilon$
0%
$a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon$

Explanation

It is fundamental concept that, for limit of a function

$f(x)$ to exist at any point

$a$ there exists a real number

$\delta>0$ , such that

$0< |x-a|<\delta$ , for which

$|f(x)-L| < \epsilon$ , where

$\epsilon >0$

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

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

Explanation

$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$

$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$

$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) }$

$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$

$\displaystyle \lim_{x\rightarrow 0}\frac{(1-e^{x})\sin x}{x^{2}+x^{3}}=$

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

Explanation

The expression can be simplified as,
(lim

$x \rightarrow 0 \dfrac{(1-{e}^{x})}{x}) \times (\dfrac{sin x}{x})( lim _{ x \rightarrow 0} \dfrac{1}{1+x} )$
Now using,
lim

$x \rightarrow 0 \dfrac{({e}^{x} - 1)}{x} = 1$
lim

$x \rightarrow 0 \dfrac{sin x}{x}$

$= 1$
We get,

$(-1)$

$\times 1 \times lim x \rightarrow 0 \dfrac{1}{1+x}$

$= -1$

$f(x)=2x-3,a=2,l=1$ and

$\epsilon =0.001$ then

$\delta>0$ satisfying

$0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon$ , is:

0%
$0.005$
0%
$0.0005$
0%
$0.001$
0%
$0.0001$

Explanation

$|f(x)-l|<0.001=\epsilon$

$\Rightarrow |2x-3-1| < 0.001$

$\Rightarrow -0.001<2x-4<0.001$

$\Rightarrow -0.0005<x-2<0.0005$

$\Rightarrow |x-2|<0.0005$

$\Rightarrow |x-a|<0.0005=\delta$

Hence,

$\delta = 0.0005 > 0$

$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$

0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$\dfrac{3}{2}$

Explanation

We simplify the given expression as,

$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)}$

Let

$x - sin x = y$

$x \rightarrow 0 \text{ so does y } \rightarrow 0$

Hence, the question transforms into

$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y})$

$= \dfrac{1}{2} \times 1$

$= \dfrac{1}{2}$

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

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

Explanation

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

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

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } }$

$=\displaystyle \dfrac { 1 }{ 2 }$

$\displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)$ is finite then the value of

$a,b$ respectively are

0%
$5\, -4$
0%
$-5,\ -4$
0%
$-4,\ 3$
0%
$4,\ 5$

Explanation

$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)$

${ x\rightarrow 0 }$ , denominator tends to 0, so the numerator also tends to 0.

$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$

$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$

$\Rightarrow 1+a+b=0$

$\Rightarrow a+b=-1$

Now, put

$a = -4$

$\Rightarrow b = 3$

Option C satisfies above equation.

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

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{3}{2}$
0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{1}{4}$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}$

$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$ ...... As

$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$ =

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

Explanation

$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } }$

$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } }$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1$

Solve :

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$

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

Explanation

Solve :

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$

$=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)$

$=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}$

$=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}$

$=\dfrac34$

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

0%
$\dfrac{10}{3}$
0%
$\dfrac{3}{10}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{5}{6}$

Explanation

Using,

$1 - cos 2x = 2{sin}^{2} x$ ,
The expression transforms to,

$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x}$
Rewriting the expression in a different form,

$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3}$
Therefore, the limit to the expression is

$\dfrac{10}{3}$
Hence, option 'A' is correct.

The value of

$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] }$ is:

0%
$0$
0%
$1$
0%
$\displaystyle \frac { \sin { \beta } }{ \beta }$
0%
$\displaystyle \frac { \sin { 2\beta } }{ 2\beta }$

Explanation

$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] }$

$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] }$

$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right)$ .......

$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$

$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$ .....

$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$

$=\dfrac { \sin { 2\beta } }{ 2\beta }$

Evaluate:

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

0%
$\displaystyle \frac{3}{2}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{3}{4}$

Explanation

Given,

$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

$\sec x=\dfrac{1}{\cos x}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$

$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$

$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$

$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$

$\cos 0=1$ and

$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$

$=\dfrac{3x}{2x}=\dfrac{3}{2}$

$\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}$ =

0%
$1$
0%
$\displaystyle \frac{2}{\pi}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{5}$

Explanation

$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x }$

$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } }$

$\displaystyle =\dfrac{2}{\pi}$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$ =

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

Explanation

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$

$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }$

$=\sqrt { 2 }$

$\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$

0%
$\sqrt{3}$
0%
$\dfrac{1}{\sqrt{3}}$
0%
$-\sqrt{3}$
0%
$-\dfrac{1}{\sqrt{3}}$

Explanation

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }$

$=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }$

$=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}$

$\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=$

0%
0
0%
1
0%
$\displaystyle \frac{\pi}{180}$
0%
$\pi$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$

$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x }$

$\displaystyle={ \dfrac { \pi }{ 180 } }$

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$ =

0%
$0$
0%
$1$
0%
$(\displaystyle \frac{\pi}{180})^{3}$
0%
$4.(\displaystyle \frac{\pi}{180})^{3}$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$

$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }$

$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } }$

$\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } }$

$\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }$

Solve:

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}$

0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{4}$
0%
$4$
0%
$-4$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3\tan x-\tan 3x}{2x^{3}}$

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+......)-(3x+\dfrac{(3x)^3}{3}+\dfrac{2(3x)^5}{15}+....)}{2x^{3}}=\dfrac{1-9}{2}=-4$

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$ =

0%
$\displaystyle \log(\frac{a}{b})$
0%
$\displaystyle \log(\frac{b}{a})$
0%
$\log(ab)$
0%
${\log_{b}}{a}$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$

$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-1-b^{x}+1}{{x}}$

$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(a^{x}-1)-(b^{x}-1)}{{x}}$

$=\log a - \log b$ ......... Using formula

$=\log\left(\dfrac{a}{b}\right)$

Hence,

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}=\log\left(\dfrac{a}{b}\right)$

$\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})$ is:

0%
does not exist
0%
is equal to 0
0%
is equal to 1
0%
exists and different from 0 and 1

Explanation

$sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)$

$sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$

$\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)$
sin

$(\dfrac{1}{x})$ is an oscillatory function means same finite no.

$0\times finite\ no.=0$

$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$ =

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

Explanation

$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$

$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}$ .................(dividing the num and deno by cosx)

$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1$

$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$

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

Explanation

$Put\ \theta = \dfrac{\pi }{2}-\phi$

$\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$ =

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{1}{8}$

Explanation

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } }$ ......(x-->h-pi/2)

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }$

$\displaystyle =\dfrac { 1 }{ 8 }$

$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$

0%
$1$
0%
$-1$
0%
$\dfrac{1}{3}$
0%
$0$

Explanation

$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}$

$=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}$

$=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1$

$\displaystyle f(x)=\left\{ \begin{matrix} \dfrac { 3 }{ { x }^{ 2 } } \sin { 2{ x }^{ 2 } } ,\; x<0 \\ \dfrac { { x }^{ 2 }+2x+C }{ 1-3{ x }^{ 2 } } ,\; x\ge 0,\; x\neq \frac { 1 }{ \sqrt { 3 } } \\ 0,\; x=\frac { 1 }{ \sqrt { 3 } } \end{matrix} \right.$

$\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=$

0%
$\frac{C}{4}$
0%
2C
0%
$\frac{C}{3}$
0%
C

Let

$f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases},$

The quadratic equation whose roots are

$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)$ and

$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)$ is

0%
$x^{2}-10x+21=0$
0%
$x^{2}-6x+9=0$
0%
$x^{2}-14x+49=0$
0%
$x^{2}+6x+9=0$

Explanation

For

$0 < x < 2$ ,

$f(x) = {x}^{2} - 1$

$\lim_{ x \rightarrow {2}^{-}} f(x) = 3$
For,

$2 < x < 3$

$f(x) = 2x + 3$

$\lim_{ x \rightarrow {2}^{+}} f(x) = 7$
Thus the equation with roots, 3 and 7 is

${x}^{2} - 10x + 21 = 0$

The value of

$\sqrt{e}$ e upto four decimal places is?

0%
$1.6484$
0%
$1.5237$
0%
$1.2589$
0%
$1.9190$

Explanation

$(e)^{\frac {1}{2}}=1+\dfrac {1}{2}+\dfrac {(\dfrac {1}{2})^2}{2!}+\dfrac {(\dfrac {1}{2})^3}{3!}+\dfrac {(\dfrac {1}{2})^4}{4!}$

$=1+0\cdot 5+0\cdot 125+0\cdot 02083+0\cdot 00260$

$=1\cdot 6484$

$f(x) = \displaystyle \frac {x^2+6x}{\sin x}$ , then

$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=$

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\lim_{x\rightarrow 0 }^{ - }f\left( x \right) =\dfrac { { x }^{ 2 }+6x }{ \sin x } \\= \lim_{ h\rightarrow 0 } f\left( h \right)$

$=\dfrac { { h }^{ 2 }+6(0-h) }{ \sin(0-h) }= \lim_{h\rightarrow 0 }f\left( h \right) =\dfrac { { h }-6 }{ -\sin(h)/h } = \dfrac { -6 }{ -1 } =6$
Hence, the answer is "C".

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$ =

0%
$a+b$
0%
$a-b$
0%
$-(a+b)$
0%
$ab$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$

$=\displaystyle a.\lim_{x\to 0}\frac{\log(1+ax)}{ax}-b.\lim_{x\to 0}\frac{\log(1+bx)}{bx}=a-b$

Let

$f$ be a continuous function on [1,3]. lf

$f$ takes only rational values for all

$x$ and

$f(2)=10$ then

$f(1.5)$ is equal to

0%
$8$
0%
$\displaystyle \frac{f(1)+f(3)}{3}$
0%
$20$
0%
$10$

Explanation

Since

$f$ is continuous, so it must take all real values between

$f\left(1\right)$ and

$f\left(3\right)$ .

But since

$f$ takes only rational values so

$f$ must be a constant function.

Hence

$f\left(1.5\right)=f\left(2\right)=10$

Evaluate:

$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$

0%
$0$
0%
$-1$
0%
$4$
0%
$-6$

Explanation

Given,

$\lim _{x\to \:0}\left(\dfrac{\sin \left(3\right)x^2}{\cos \left(2x^2-x\right)}\right)$

$=\dfrac{\sin \left(3\right)\cdot \:0^2}{\cos \left(2\cdot \:0^2-0\right)}$

$=0$

$\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { 3\sin(2{ x }^{ 2 }) }{ { x }^{ 2 } } } =A$

then the value of

$A$ is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\displaystyle \lim_{x\rightarrow { 0^- }} \dfrac { 3\sin({ 2x }^{ 2 }) }{ { x }^{ 2 } } =\lim_{x\rightarrow { 0^- }}\dfrac { 2*3\sin({ 2x }^{ 2 }) }{ { 2x }^{ 2 } } = 6$

Hence, answer is option

$C$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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