If $$f(x)=\displaystyle \frac { 1 }{ x(3x+1) } $$ then at $${x}=0,\mathrm{f}({x})$$ is

$$\displaystyle \lim_{x\rightarrow 0 }f(x)=2$$

MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 4

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1+a^{3})+8e^{1/x}}{1+(1-b^{3})e^{1/x}}=2$ then

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$a=1,b=(-3)^{1/3}$
0%
$a=1,b=3^{1/3}$
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$a=-1,b=(-3)^{1/3}$
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$a=1,b=0$

Explanation

We have,

$\displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + (1 - b^3) e^{1/x}} = 2$

$\displaystyle \left [ \frac{\infty}{\infty} form \right ]$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) e^{-1/x} + 8}{e^{-1/x} + (1 - b^3)} = 2$

$\Rightarrow \displaystyle \frac{0 + 8}{0 + (1 - b^3)} = 2 \Rightarrow 1 - b^3 = 4 \Rightarrow b^3 = - 3 \Rightarrow b = (-3)^{1/3}$

Again,

$\displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + (1 - b^3) e^{1/x}} = 2$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + 4e^{1/x}} = 2$

$\Rightarrow 1 + a^3 = 2$ [On comparison]

$\Rightarrow a = 1$
Therefore,

$a = 1, b = (-3)^{1/3}$
Hence, option 'A' is correct.

$f(x)=(x)^{\tfrac{1}{x-1}}$ for

$x\neq 1$ and

$\mathrm{f}$ is continuous at

$\mathrm{x}=1$ then

$\mathrm{f}(1)=$

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

Explanation

Given that the function is continuous at x= 1

$\Rightarrow \displaystyle \lim _{ x\rightarrow 1 }{ f(x) } = f(1)$

$\displaystyle \lim _{ x\rightarrow 1 }{ f(x) } =\quad \lim _{ x\rightarrow 1 }{ {\ x }^{\ { 1 }/\left( { x-1 } \right) } } \\ \displaystyle =\lim _{ x\rightarrow 1 }{ { \left( 1+\quad x-1 \right) }^{ { 1 }/{ \left( x-1 \right) } } } \left( { 1 }^{ \infty }\quad form \right) \\ \displaystyle ={ e }^{\displaystyle \lim _{ x\rightarrow 1 }{ \frac { x-1 }{ x-1 } } }\\ =e$
Hence f(1) is also equal to e.

Assertion (A):

$f(x)=\displaystyle \frac{\sin\{[x]\pi\}}{1+x^{2}}$ is continuous on

$\mathrm{R}$ (where

$[x]$ denotes greatest integer function of

$x$ ).
Reason (R): Every constant function is continuous on

$\mathrm{R}$

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true and R is not the correct explanation of A
0%
A is true but R is false
0%
R is true but A is false

Explanation

$f\left( x \right) =\cfrac { \sin { \left\{ \left[ 2 \right] \pi \right\} } }{ 1+{ x }^{ 2 } }$

$\left[ x \right]$ will always give us an integer

and we know that

$\sin { \left( n\pi \right) } =0,n\epsilon I$

$\Rightarrow f\left( x \right) =\cfrac { 0 }{ 1+{ x }^{ 2 } } =0$

$\Rightarrow f\left( x \right)$ is a constant function.

$\Rightarrow f\left( x \right)$ is continuous on R.

The values of

$p$ and

$q$ for which the function

$\mathrm{f}(\mathrm{x}) = \left\{\begin{array}{ll} \dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} & , \mathrm{x}<0\\ \mathrm{q} & , \mathrm{x}=0\\ \dfrac{\sqrt{\mathrm{x}+\mathrm{x}^{2}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{3/2}} & , \mathrm{x}>0 \end{array}\right.$
is continuous for all

$\mathrm{x}$ in

$\mathrm{R}$ , are

0%
$\displaystyle \mathrm{p}=\dfrac{1}{2},\ \displaystyle \mathrm{q}=-\dfrac{3}{2}$
0%
$\displaystyle \mathrm{p}=\dfrac{5}{2},\mathrm{q}=\dfrac{1}{2}$
0%
$\displaystyle \mathrm{p}=-\dfrac{3}{2},\ \displaystyle \mathrm{q}=\dfrac{1}{2}$
0%
$\displaystyle \mathrm{p}=\dfrac{1}{2},\mathrm{q}=\dfrac{3}{2}$

Explanation

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll} \dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} & , \mathrm{x}<0\\ \mathrm{q} & , \mathrm{x}=0\\ \dfrac{\sqrt{\mathrm{x}+\mathrm{x}^{2}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{3/2}} & , \mathrm{x}>0 \end{array}\right.$

$\displaystyle \lim_{\mathrm{x}\rightarrow 0}\mathrm{f}(\mathrm{x})=\lim_{\mathrm{x}\rightarrow 0^{+}}\dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}}=\mathrm{p}+2$

$\displaystyle \lim_{\mathrm{x}\rightarrow 0^{-}}\mathrm{f}(\mathrm{x})=\dfrac{1}{2}= \displaystyle \mathrm{p}+2=\mathrm{q}=\dfrac{1}{2}$

$p+2=\dfrac12\, , q=\dfrac12$

$\therefore \displaystyle \mathrm{p}=-\dfrac{3}{2},\ \displaystyle \mathrm{q}=\dfrac{1}{2}$
Hence, option 'C' is correct.

$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} x+\lambda , & -1<x<3 \end{matrix} \\ \begin{matrix} 4, & x=3 \end{matrix} \\ \begin{matrix} 3x-5, & x>3 \end{matrix} \end{cases}$ is continuous at

$x=3$ then tha value of

$\lambda$ is

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$-1$
0%
$0$
0%
$1$
0%
does not exits

Explanation

Function

$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} x+\lambda , & -1>x>3 \end{matrix} \\ \begin{matrix} 4, & x=3 \end{matrix} \\ \begin{matrix} 3x-5, & x>3 \end{matrix} \end{cases}$ is continuous at

$x=3$ .

We know that as

$f(x)$ is continuous at

$x=a,$ therefore

$\displaystyle\lim _{ x\rightarrow a }{ f\left( x \right) } =f\left( a \right)$ .

Since the given function is continuous at

$x=3,$

therefore

$\displaystyle \lim _{ x\rightarrow 3- }{ \left( x+\lambda \right) } =f\left( 3 \right) \Rightarrow 3+\lambda =4\Rightarrow \lambda =4-3=1$

The function

$\displaystyle \mathrm{f}({x})=\begin{cases}\dfrac{1-\sin x}{(\pi-2x)^{2}} & x \neq\dfrac{\pi}{2}\\ \mathrm{k}& {x}=\dfrac{\pi}{2}\end{cases}$ is continuous at

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

$\mathrm{k}$ is equal to

0%
$\dfrac{1}{8}$
0%
4
0%
3
0%
1

Explanation

Substitute

$x=\dfrac{\pi }{2}-t$ ,

$\lim _{t\rightarrow 0} \dfrac{1-sin(\dfrac{\pi }{2}-t)}{(2t)^{2}}$

Applying L-Hospital's rule twice ,we get

$=\dfrac{1-cos t}{4 t^{2}}=\dfrac{1}{8}$

So for continuity

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

The value of

$f(0)$ so that the function

$\mathrm{f}({x})$

$=\displaystyle \frac{1-\cos(1-\cos x)}{x^{4}}$ is continuous everywhere is

0%
$\displaystyle \frac{1}{8}$
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$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{3}$

Explanation

$f\left( x \right) =\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) }$

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { x }^{ 4 } } }$

$=\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times \cfrac { { (1-\cos { x } ) }^{ 2 } }{ { x }^{ 4 } }$

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times \displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { 1-\cos { x } }{ { x }^{ 2 } } \right) ^{ 2 } } }$

$x\rightarrow 0$

$1-\cos { x } \rightarrow 0$

$=\displaystyle \lim _{ 1-\cos { x } \rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }$

$=\cfrac { 1 }{ 2 } \times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }=\cfrac { 1 }{ 8 }$

$\underset{x\rightarrow0}{lim}\displaystyle\frac{1-cos^{3}x+sin^{3}x+\ell n(1+x^{3})+\ell n(1+cos\,\,x)}{x^{2}-1+2\,cos^{2}x+tan^{4}x+sin^{3}x}$ is equal to -

0%
$\,\,\displaystyle\frac{3}{4}$
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$\,\,ln2$
0%
$\,\,\displaystyle\frac{ln2}{4}$
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$\,\,3/2$

Explanation

$\displaystyle \underset { x\rightarrow 0 }{ lim } \frac { 1-cos^{ 3 }x+sin^{ 3 }x+\ell n(1+x^{ 3 })+\ell n(1+cos\, \, x) }{ x^{ 2 }-1+2\, cos^{ 2 }x+tan^{ 4 }x+sin^{ 3 }x }$

$\displaystyle =\frac { 1-1+0+\ln { \left( 1+0 \right) } +\ln { \left( 1+1 \right) } }{ 0-1+2+0+0 } =\ln { 2 }$

The value of

$p$ for which the function

$f(x)=\displaystyle \left\{ \begin{array}{rl} \dfrac { (4^{ { x } }-1)^{ 3 } }{ \sin\dfrac { { x } }{ { p } } \log(1+\dfrac { { x }^{ 2 } }{ 3 } ) } & { ;\quad x\neq 0 } \\ 12(\log 4)^{ 3 } & { ;\quad x=0\quad } \end{array} \right.$ is continuous at

${x}=0$ , is

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

Explanation

We know that the function is continuous at

$x=0$ if

$\displaystyle \lim _{ x\rightarrow0 }{ f(x) }$ exists and is equal to

$f(0)$ .

Therefore, for continuity we have,

$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { 4 }^{ x }-1 \right) }^{ 3 } }{ \sin { \dfrac { x }{ p } } \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 } \right) } } } =f\left( 0 \right) =12{ \left( \log { 4 } \right) }^{ 3 }$

Since

$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { a }^{ x }-1 \right) } }{ x } } =\log { a }$ for

$a\neq 0,a>1$ ,

$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \sin { x } }{ x } =1 }$ , and

$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \log { \left( 1+x \right) } }{ x } =1 }$ ,

the limit on LHS can be rewritten as:

$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \dfrac { { 4 }^{ x }-1 }{ x } \right) }^{ 3 }\cdot { x }^{ 3 } }{ \dfrac { \sin { \dfrac { x }{ p } } }{ \dfrac { x }{ p } } \cdot \dfrac { x }{ p } \cdot \dfrac { \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 } \right) } }{ \dfrac { { x }^{ 2 } }{ 3 } } \cdot \dfrac { { x }^{ 2 } }{ 3 } } }$

$=\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \log { 4 } \right) }^{ 3 }\cdot { x }^{ 3 } }{ 1\cdot \dfrac { x }{ p } \cdot 1\cdot \dfrac { { x }^{ 2 } }{ 3 } } }$ (using standard limits stated above)

$=3p{ \left( \log { 4 } \right) }^{ 3 }$ .... (because

$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { { x }^{ 3 } }{ { x }^{ 3 } } =1 }$ )

According to the condition for continuity, this must equal

$f\left( 0 \right) =12{ \left( \log { 4 } \right) }^{ 3 }$

$3p{ \left( \log { 4 } \right) }^{ 3 }=12{ \left( \log { 4 } \right) }^{ 3 }$

$\\ \Rightarrow 3p=12$

$\\ \Rightarrow p = 4$

$f(x)=\displaystyle \frac { 1 }{ x(3x+1) }$ then at

${x}=0,\mathrm{f}({x})$ is

0%
continuous
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discontinuous
0%
not determined
0%
$\displaystyle \lim_{x\rightarrow 0 }f(x)=2$

Explanation

$f\left( x \right) =\cfrac { 1 }{ x\left( 3x+1 \right) }$

LHL

$=\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \cfrac { 1 }{ x\left( 3x+1 \right) } }$

$x\rightarrow { 0 }^{ - },\cfrac { 1 }{ x } \rightarrow -\infty$

$=\cfrac { -\infty }{ 0+1 } =-\infty$

RHL

$=\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x(3x+1) } }$

$x\rightarrow { 0 }^{ + },\cfrac { 1 }{ x } \rightarrow +\infty$

$=\cfrac { +\infty }{ 0+1 } =\infty$

LHL

$\neq$ RHL

$\Rightarrow$ Discontinuous

Assertion(A):

$f(x)=x(\displaystyle \frac{1+e^{1/x}}{1-e^{1/x}})(x\neq 0)$ ,

${f}(0)=0$ is continuous at

${x}=0$ .
Reason(R) A function is said to be continuous at

$a$ if both limits are exists and equal to

$\mathrm{f}({a})$ .

0%
Both A and R are true and R is the correct explanation of A
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Both A and R are true and R is not the correct explanation of A
0%
A is true but R is false
0%
R is true but A is false

Explanation

Assertion :

$f(x)=x(\displaystyle \frac{1+e^{1/x}}{1-e^{1/x}})(x\neq 0)$

$LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ x\left( \dfrac { 1+e^{ 1/x } }{ 1-e^{ 1/x } } \right) } =\lim _{ h\rightarrow 0 } (-h)\left( \displaystyle\frac { 1+e^{ -1/h } }{ 1-e^{ -1/h } } \right) =0$

$RHL=\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ x\left( \dfrac { 1+e^{ 1/x } }{ 1-e^{ 1/x } } \right) }$

$=\lim _{ h\rightarrow 0 } h\left( \dfrac { 1+e^{ 1/h } }{ 1-e^{ 1/h } } \right) =\lim _{ h\rightarrow 0 } h\left( \displaystyle\frac { e^{ -1/h }+1 }{ e^{ -1/h }-1 } \right) =0$

Given

$f(0)=0$
So,

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

$f(x)$ is continuous at

$x=0$

Also, reason is true.

The value of

$\mathrm{f}({0})$ so that the function

$\displaystyle \mathrm{f}({x})=\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$ becomes continuous is equal to

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{4}$
0%
$2$
0%
$\dfrac{1}{3}$

Explanation

Since, f is continuous

$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { \sqrt { 1+x } -\sqrt [ 3 ]{ 1+x } }{ x } }$

It is of

$\dfrac{0}{0}$ form

By L-Hospital's rule

$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \left( \frac { 1 }{ 2\sqrt { 1+x } } -\frac { 1 }{ 3{ \left( 1+x \right) }^{ 2/3 } } \right) } =\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 }$

$\displaystyle f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}}$ then

$\mathop {\lim }\limits_{x \to \infty } f(x)$ is

0%
O
0%
$\infty$
0%
1
0%
None of these

Explanation

$\displaystyle f(x) =\sqrt{\frac{x-sin x}{x+ cos^2x}} =\sqrt{\frac{1- \frac{sin x}{x}}{1+ \frac{cos^2x}{x}}}$

$\lim_{x \rightarrow \infty}f(x)=1$

$\displaystyle \lim_{x \rightarrow \infty} \frac{sin x}{x}=0$

$\displaystyle \lim_{x \rightarrow \infty} \frac{cos^2 x}{x}=0$

$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$ is

0%
$\ln (n!)$
0%
$n$
0%
$n!$
0%
$0$

Explanation

Consider that

$y=\displaystyle\lim_{x\to\infty}(1^x+2^x+3^x+.........+n^x)^{1/x}$

take natural logarithm on both sides, we have

$\ln y=\displaystyle\lim_{x\to\infty}\ln\left[(1^x+2^x+3^x+.........+n^x)^{1/x}\right]$

$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{\ln (1^x+2^x+3^x+.........+n^x)}{x}$

RHS has

$\dfrac{\infty}{\infty}$ form. Hence, use L'hospital's rule

$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n }{1^x+2^x+3^x+.........+n^x}$

On approximating this form, we have

$\ln y=\displaystyle\lim_{x\to\infty}\dfrac{n^x\ln n }{n^x}$

$\ln y=\ln n$

$\therefore y=n$

Hence, this is required answer.

$\displaystyle \lim_{x\to\infty}{\displaystyle \frac{(2x + 1)^{40}(4x - 1)^5}{(2x + 3)^{45}}}$ is equal to

0%
16
0%
24
0%
32
0%
8

Explanation

$\displaystyle\lim_{x\to\infty}\frac{(2x+1)^{40}(4x-1)^5 }{(2x+3)^{45}}$

$=\displaystyle\lim_{x\to\infty}\frac{\left(2+\Large \frac{1}{x}\right)^{40}\left(4-\Large \frac{1}{x}\right)^5}{\left(2+\Large \frac{3}{x}\right)^{45}}$ $=\displaystyle \frac{2^{40}4^5}{2^{45}}\\$

$=2^5 = 32$

$f(x) =\left\{\begin{matrix} \dfrac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}},x>0 & \\ e^{x}\sin x+\pi x+\lambda \ln 4,x\leq 0 & \end{matrix}\right.$ is continuous at

$x= 0$ , then

$\lambda$ is a

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rational number
0%
irrational number
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natural number
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complex number

Explanation

$\displaystyle\lim_{x\to-h} f(x) =\lambda \ln4$

$\displaystyle\lim_{x\to 0+h} f(x)= \dfrac { 2^{ 3x }-2^{ 2x }-2^{ x }+1^{ x } }{ x^{ 2 } } =\lambda \ln4$

$\displaystyle\Rightarrow \lim_{h\to 0}\dfrac { 2^{ 3h }-2^{ 2h }-2^{ h }+1^{ h } }{ h^{ 2 } } =\lambda \ln4$

$\displaystyle\Rightarrow \lim_{ h\to 0}\dfrac{(2^{ 2h }(2^{ h }-1)-1(2^{ h }-1))}{h^{ 2 }}=\lambda \ln4$

$\displaystyle\Rightarrow \lim_{h\to 0}\dfrac{(2^{ h }+1){ (2^{ h }-1) }^{ 2 }}{h^{ 2} }=\lambda \ln4$

$\displaystyle\Rightarrow \lim_{ h\to 0} \dfrac{2{ (2^{ h }-1) }^{ 2 }}{h^{ 2 }}=\lambda \ln4$

Applying L'hospital's rule

$\displaystyle\Rightarrow \lim_{h\to 0}(2^{ h }-1)\dfrac{\ln4}{h} =\lambda \ln4$

$\Rightarrow \ln2\ln4 = \lambda \ln4$

$\Rightarrow \ln2=\lambda$

Hence, option 'B' is correct.

$\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n + 1)^2}{(n + 2)(n^2 + 3n - 1)}}$ is equal to

0%
$0$
0%
$2$
0%
$4$
0%
$\infty$

Explanation

$\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n+1)^2}{(n+2)(n^2+3n-1)} }$

$= \displaystyle \lim_{n\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{n} \right)^2}{\left(1+\Large \frac{2}{n} \right)\left(1+\Large \frac{3}{n} - \Large \frac{1}{n^2} \right)} }$

$= \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)} = 4$

Let

$\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$ be defined by

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{k}-2\mathrm{x}, \mathrm{i}\mathrm{f} \mathrm{x}\leq-1\\ 2\mathrm{x}+3, \mathrm{i}\mathrm{f} \mathrm{x}>-1 \end{array}\right.$ be continous. then find possible value of

$\mathrm{k}$ is

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

Explanation

$\mathrm{f}(\mathrm{x})=\mathrm{k}-2\mathrm{x}$ if

$\mathrm{x}\leq-1$

$=2\mathrm{x}+3$ if

$\mathrm{x}>-1$

So,

$L.H.L.= R.H.L.$

$k-2x=2x+3$

$k=4x+3$

For

$x=-1$ ,

$k=-1$

Hence, option

$'C'$ is correct.

$\lim_{n\rightarrow \infty }\left ( \frac{F\left ( n \right )}{G\left ( n \right )} \right )^{n}-\lim_{n\rightarrow \infty }\left ( \frac{f\left ( n \right )}{g\left ( n \right )} \right )^{n}=$

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

Explanation

$e^{-3/2}-e^{-1}$ =

$f(x) = \displaystyle \left\{\begin{matrix}x - 1, & x \geq 1 \\ 2x^2 - 2, & x < 1\end{matrix}\right. , g(x) = \left\{\begin{matrix}x + 1, & x > 0 \\ -x^2 + 1, & x \leq 0\end{matrix}\right.$ , and

$h(x) = |x|$ , then

$\displaystyle \lim_{x \rightarrow 0} f(g (h (x)))$ is

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

Explanation

$\displaystyle \lim_{x \rightarrow 0^+} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0$

$\displaystyle \lim_{x \rightarrow 0^-} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0$
Therefore,

$\displaystyle \lim_{x \rightarrow 0} f(g(h(x))) = 0$
Hence, option 'A' is correct.

$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$ is non-zero finite, then

$n$ must be equal

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

Explanation

$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$

$\mbox{For n = 0,we have }\displaystyle\lim_{x\to0}\displaystyle\frac{1-1}{x-1}=0$

$\mbox{For n = 1,}\displaystyle\lim_{x\to0}\displaystyle\frac{x-\sin x}{x- \sin x}=1\\$

$\mbox{For n = 2,}\displaystyle\lim_{x\to0}\displaystyle\frac{x^2-\sin^2x}{x-\sin^2x}=\displaystyle\lim_{x\to0}\displaystyle\frac{1-\displaystyle\frac{\sin^2x}{x^2}}{\displaystyle\frac{1}{x}-\displaystyle\frac{\sin^2x}{x^2}}.\\$

$\mbox{This does not exist.}\\$

$\displaystyle { For\ \ n=3, }\lim _{ x\to 0 } \dfrac { x^{ 3 }-\sin ^{ 3 } x }{ x-\sin ^{ 3 } x } =\lim _{ x\to 0 } \dfrac { 1-\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } }{ \dfrac { 1 }{ x^{ 2 } } -\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } }$
For

$n = 3$ , also given limit does not exist.
Hence,

$n = 1$

$\displaystyle \lim_{x\to1}{\displaystyle \frac{1-x^2}{\sin 2\pi x}}$ is equal to

0%
$\displaystyle \frac{1}{2\pi}$
0%
$\displaystyle \frac{-1}{\pi}$
0%
$\displaystyle \frac{-2}{\pi}$
0%
None of these

Explanation

$\displaystyle\lim_{x\to1}\frac{1-x^2}{\sin2\pi x}= -\displaystyle \lim_{x\to1}\frac{2\pi(1-x)(1+x)}{-2\pi\sin(2\pi-2\pi x)}$

$=\displaystyle \lim_{x \to 1}\left( \frac{(2\pi-2\pi x}{\sin(2\pi-2\pi x)}\right).\frac{1+x}{-2\pi}=-\frac{1}{\pi}$

$\displaystyle \lim_{n \rightarrow \infty} \frac{-3n + (-1)^n}{4n - (-1)^n}$ is equal to

0%
$\displaystyle \frac{3}{4}$
0%
$0$ if $n$ is even
0%
$\displaystyle- \frac{3}{4}$
0%
none of these

Explanation

$\displaystyle \lim_{n \rightarrow \infty} \dfrac{-3n + (-1)^n}{4n - (-1)^n}$

$\displaystyle =\lim_{n \rightarrow \infty} \dfrac{-3 + \dfrac{(-1)^n}{n}}{4 + \dfrac{(-1)^n}{n}} = \dfrac{-3}{4}$

$\displaystyle \lim _{ x\to\infty }\left\{\displaystyle \frac{x^3 + 1}{x^2 +1} - (ax + b) \right\} = 2$ , then

0%
$a = 1, b = 1$
0%
$a = 1, b = 2$
0%
$a = 1, b = -2$
0%
None of these

Explanation

$\displaystyle\lim_{x\to\infty}\left(\frac{x^3+1}{x^2+1}-(ax+b)\right)= 2\\$

$or\displaystyle\lim_{x\to\infty}\frac{x^3(1-a)-bx^2-ax+(1-b)}{x^2+1} = 2\\$

As we have finite limit so degree of $x$ in the numerator must be less than equal to the degree of $x$ in the denominator.

So the term containing $x^3$ must have coefficient $0$ i.e $(1-a) = 0$

Hence $a =1$ .Divide Numerator and denominator by $x^3$ to get the value of $b$

$-b = 2$

$\mbox{a = 1,b = -2}\\$

Hence, option 'C' is correct.

$\displaystyle f(x) = \left\{\begin{matrix}x^2+2, & x \geq 2\\ 1-x, & x < 2\end{matrix}\right.$ and

$g(x) = \left\{\begin{matrix}2x, & x > 1\\ 3-x, & x \leq 1\end{matrix}\right.$ , then the value of

$\displaystyle \lim_{x \rightarrow 1} f(g(x))$ is ............

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

Explanation

$\displaystyle \lim_{x \rightarrow 1^+} f(g(x)) = f(g(1^+)) = f((2^+)) = 2^2 + 2 = 6$
and

$\displaystyle \lim_{x - 1^-} f(g(x)) = f(g(1^-)) = f(3 - 1^-) =f(2^+) = 2^2 + 2 = 6$
Hence,

$\displaystyle \lim_{x \rightarrow 1} f(g(x)) = 6$ .

$\displaystyle \lim_{x\to2} \left( \left( \displaystyle \frac{x^3 - 4x}{x^3 - 8} \right)^{-1} - \left( \displaystyle \frac{x + \sqrt{2x}}{x - 2} - \displaystyle \frac{\sqrt {2}}{\sqrt{x} - \sqrt{2}} \right)^{-1} \right)$ is equal to

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

Explanation

$\displaystyle \lim _{ x\rightarrow 2 }{ \left( { \left( \cfrac { x\left( { x }^{ 2 }-4 \right) }{ { x }^{ 3 }-8 } \right) }^{ -1 }-{ \left( \cfrac { \left( \sqrt { x } +\sqrt { 2 } \right) \sqrt { x } }{ { \left( \sqrt { x } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 } } -\cfrac { \sqrt { 2 } }{ \sqrt { x } -\sqrt { 2 } } \right) }^{ -1 } \right) }$

$=\displaystyle \lim _{ x\rightarrow 2 }{ \left( { \left( \cfrac { x(x-2)(x+2) }{ (x-2)({ x }^{ 2 }+2x+4) } \right) }^{ -1 }-{ \left( \cfrac { \sqrt { x } }{ \sqrt { x } -\sqrt { 2 } } -\cfrac { \sqrt { 2 } }{ \sqrt { x } -\sqrt { 2 } } \right) }^{ -1 } \right) }$

$=\displaystyle \lim _{ x\rightarrow 2 }{ \left( \left( \cfrac { { x }^{ 2 }+2x+4 }{ { x }^{ 2 }+2x } \right) -1 \right) }$

$=\displaystyle \lim _{ x\rightarrow 2 }{ \left( 1+\cfrac { 4 }{ { x }^{ 2 }+2x } -1 \right) }$

$=\cfrac { 4 }{ { 2 }^{ 2 }+2\cdot 2 } =\cfrac { 4 }{ 8 } =\cfrac { 1 }{ 2 }$

$\displaystyle f(x) = \frac{3x^2 + ax + a + 1}{x^2 + x - 2}$ and

$\displaystyle \lim_{x \rightarrow - 2} f(x)$ exists.

Then the value of

$(a- 4)$ is?

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

$\displaystyle f(x) = \frac{3x^2 + ax + a + 1}{(x + 2)(x-1)}$
As

$x \rightarrow -2, D^r \rightarrow 0$ . Hence, as

$x \rightarrow -2, N^r \rightarrow 0$ . Therefore,

$\therefore 12 - 2a + a + 1 = 0$ or

$a = 13$
Hence, option 'A' is correct.

$\displaystyle \lim_{n \rightarrow \infty} {^{n}C_{c}}\left(\dfrac {m}{n}\right)^{x}\left(1-\dfrac {m}{n}\right)^{n-x}$ equal to

0%
$\dfrac {M^{x}}{x\ !}.e^{-m}$
0%
$\dfrac {M^{x}}{x\ !}.e^{m}$
0%
$0$
0%
$\dfrac {m^{x+1}}{me^{m}x\ !}$

Explanation

Given

$\displaystyle\lim_{n\rightarrow \infty}{^{n}}C_x \bigg(\dfrac{m}{n}\bigg)^{x}\bigg(1-\dfrac{m}{n}\bigg)^{n-x}=$ coefficient of

$x$ in

$\displaystyle\lim_{n\rightarrow\infty}\bigg(\dfrac{m}{n}+1-\dfrac{m}{n}\bigg)^{n}$

$=$ coeeficient of

$x$ in

$\displaystyle\lim_{n\rightarrow\infty}\bigg(1\bigg)^{n}=$ coefficient of

$x$ in

$1=0$

$\displaystyle \lim_{x\to0}\left( x^{-3}\sin{3x} + ax^{-2} + b \right)$ exists and is equal to 0, then

0%
$a = -3$ and $b = \dfrac{9}{2}$
0%
$a = 3$ and $b = \dfrac{9}{2}$
0%
$a = -3$ and $b = -\dfrac{9}{2}$
0%
$a = 3$ and $b = -\dfrac{9}{2}$

Explanation

$\displaystyle\lim_{x\to\infty}\displaystyle\frac{\sin3x}{x^3}+\displaystyle\frac{a}{x^2}+b = \lim_{x\to0}\displaystyle\frac{\sin3x+ax+bx^3}{x^3}$

$=\displaystyle\lim_{x\to0}\displaystyle\frac{3\displaystyle\frac{\sin3x}{3x}+a+bx^2}{x^2}$

$\mbox{For existence, }$

$(3+a)=0 \mbox{ or } a=-3$

$\therefore L=\displaystyle\lim_{x\to0}\displaystyle\frac{\sin3x-3x+bx^3}{x^3}$

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\sin t - t}{t^3}+b = 0 ........ (3x=t )$
The left hand side reduces to

$\dfrac{0}{0}$ form by substituting the limit

Then, using L'Hospital's rule

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\cos t - 1}{3t^2}+b = 0$

Again, applying L'Hospital's rule

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{-\sin t}{6t}+b = 0$

$\Rightarrow -\displaystyle\frac{27}{6}+b=0 \mbox{ or } b=\frac{9}{2}$

$\displaystyle \lim _{ x\to\infty } \left( \frac { x^{ 2 }+2x-1 }{ 2x^2-3x-2 } \right) ^{\LARGE \frac { 2x+1 }{ 2x-1 } }$ is equal to

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

Explanation

$\displaystyle\lim_{x\to\infty}\left(\displaystyle \frac{x^2+2x-1}{2x^2-3x-2}\right)^{\displaystyle \frac{2x+1}{2x-1}}\\$

$=\displaystyle\lim_{x\to\infty}\left(\displaystyle \frac{1+\displaystyle \frac{2}{x}-\displaystyle \frac{1}{x^2}}{2-\displaystyle \frac{3}{x}-\displaystyle \frac{2}{x^3}}\right)^{\displaystyle \frac{2+{{1}/{x}}}{2-{{1}/{x}}}}\\$

$=1/2$

$p\left( x \right) ={ a }_{ 0 }+{ a }_{ 1 }x+...+{ a }_{ n }{ x }^{ n }$ and

$\left| p\left( x \right) \right| \le \left| { e }^{ x-1 }-1 \right|$ for all

$x\ge 0,$ then

$\left| { a }_{ 1 }+2{ a }_{ 2 }+3{ a }_{ 3 }+...+n{ a }_{ n } \right|$

0%
$\le 1$
0%
$\ge 1$
0%
$\ge 0$
0%
$\le 0$

Explanation

We have,

$\displaystyle p'\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \frac { p\left( 1+h \right) -p\left( 1 \right) }{ h } } \Rightarrow \left| p'\left( 1 \right) \right| =\left| \lim _{ h\rightarrow 0 }{ \frac { p\left( 1+h \right) -p\left( 1 \right) }{ h } } \right|$

$\displaystyle =\lim _{ h\rightarrow 0 }{ \frac { \left| p\left( 1+h \right) -p\left( 1 \right) \right| }{ \left| h \right| } } \le \lim _{ h\rightarrow 0 }{ \frac { \left| p\left( 1+h \right) +p\left( 1 \right) \right| }{ \left| h \right| } }$ ........(1)

Now,

$\left| p\left( x \right) \right| \le \left| { e }^{ x-1 }-1 \right| \Rightarrow \left| p\left( 1-h \right) \right| \le \left| { e }^{ h }-1 \right|$ and

$\left| p\left( 1 \right) \right| \le 0\Rightarrow p\left( 1 \right) =0$

Hence,

$\displaystyle \left| p'\left( 1 \right) \right| \le \lim _{ h\rightarrow 0 }{ \frac { \left| { e }^{ h }-1 \right| }{ \left| h \right| } } =1$

$\displaystyle \Rightarrow \left| p'\left( 1 \right) \right| \le 1\Rightarrow \left| { a }_{ 1 }+2{ a }_{ 2 }+3{ a }_{ 3 }+...+n{ a }_{ n } \right| \le 1$

The value of

$\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$ for

$(a>1)$ is equal to?

0%
$1$
0%
$0$
0%
$\displaystyle\frac{\pi}{2}$
0%
Does not exist

Explanation

$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$

$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$
Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.

The value of

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

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

Explanation

We have

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

$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{(2 sin x - 1)(sin x + 1)}{(2 sin x - 1)(sin x - 1)}$

$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{sin x + 1}{sin x - 1} = - 3$

Let

$f(x)=\sin x$ ,

$g(x)=\left [ x+1 \right ]$ and

$g(f(x))=h(x)$ , where [.] is the greatest integer function. Then

$h^+\left ( \displaystyle \dfrac{\pi }{2} \right )$ is

0%
non existent
0%
$1$
0%
$-1$
0%
none of these

Explanation

$f(x) =\sin(x)$ and

$g(x) = [1+x]$

$\Rightarrow g(f(x)) = [1+\sin x]=h(x)$

$h^{+}\left( \cfrac{\pi}{2}\right) =[1+\sin(\pi/2)]= 1$

Which one of the following statement is true?

0%
if $\displaystyle \lim_{x\to c}f(x).g(x)$ and $\displaystyle \lim_{x\to c}f(x)$ exist, then $\displaystyle \lim_{x\to c}g(x)$ exists
0%
if $\displaystyle \lim_{x\to c}f(x).g(x)$ exists, then $\displaystyle \lim_{x\to c}f(x)$ and $\displaystyle \lim_{x\to c}g(x)$ exist
0%
if $\displaystyle \lim_{x\to c}(f(x)+g(x))$ and $\displaystyle \lim_{x\to c}f(x)$ exist, then $\displaystyle \lim_{x\to c}g(x)$ exists
0%
if $\displaystyle \lim_{x\to c}(f(x)+g(x))$ exists, then $\displaystyle \lim_{x\to c}f(x)$ and $\displaystyle \lim_{x\to c}g(x)$ exists

Explanation

The statements can be proved false by giving counter examples

(1) If

$\displaystyle \lim_{x\to c}f(x).g(x)$ and

$\displaystyle \lim_{x\to c}f(x)$ exist, then

$\displaystyle \lim_{x\to c}g(x)$ exists
Let

$g(x) = 1/x$ and

$f(x) = 2x$ , with

$c = 0$ ,
then

$\displaystyle \lim_{x\to c}f(x).g(x) =2$

$\displaystyle \lim_{x\to c}f(x) = 0$ ,but $\displaystyle \lim_{x\to c}g(x)$ does not exist

(2) Same example as (1)
(4) Let

$g(x) =\dfrac{ x+ 1}{x}$ and

$f(x) = \dfrac{x-1}{x}$ , with

$c = 0$ ,
then

$\displaystyle \lim_{x\to c}[f(x)+g(x)] =2$

but $\displaystyle \lim_{x\to c}f(x) \ and \ g(x)$ do not exist

$\displaystyle \lim_{n\to\infty }\frac{n^{p}\sin ^{2}\left ( n! \right )}{n+1}$ ,

$0<p<1$ , is equal to

0%
$0$
0%
$\infty$
0%
$1$
0%
$none\ of\ these$

Explanation

We have,

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

$\Rightarrow \displaystyle \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n + 1} = \lim_{n \rightarrow \infty} \dfrac{sin^2 (n!)}{n^{1 - p}} \times \dfrac{1}{1 + \dfrac{1}{n}}$

$\displaystyle \Rightarrow \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n+1} = \dfrac{\text{An oscillating number}}{\infty} \times 1 = 0 \times 1 = 0$ .
Hence, option 'A' is correct.

$\displaystyle \lim_{x \rightarrow \infty} (\sqrt{x^2 + 8x + 3} - \sqrt{x^2 + 4x + 3}) =$

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

Explanation

Let

$x = \displaystyle \dfrac{1}{y}$ then as

$x \rightarrow \infty \Rightarrow y \rightarrow 0$

$\therefore \displaystyle \lim_{y \rightarrow 0} \left ( \sqrt{\dfrac{1}{y^2} + \dfrac{8}{y} + 3} - \sqrt{\dfrac{1}{y^2} + \dfrac{4}{y} + 3} \right )$

$\therefore \displaystyle \lim_{y \rightarrow 0}\frac{\sqrt{3y^2 + 8y + 1} - \sqrt{3y^2 + 4y + 1}}{y}$

$= \displaystyle \lim_{y \rightarrow 0} \left ( \dfrac{\displaystyle \dfrac{1}{2} (3y^2 + 8y + 1)^{-1/2} (6y + 8) - \dfrac{1}{2} (5y^2 + 4y + 1)^{-1/2} (6y + 4)}{1} \right )$

$= \displaystyle \lim_{y \rightarrow 0} \left ( \frac{8}{2} - 2 \right ) = 2$

Let

$f\left ( x \right )=\begin{cases}\sin x, x\neq n\pi \\ 2, x=n\pi \end{cases}$ , where

$n\epsilon \mathbb{Z}$ and

$g\left ( x \right )=\begin{cases}x^{2}+1, x\neq 2 \\ 3, x=2 \end{cases}$ .
Then

$\displaystyle \lim_{x\to 0}g\left ( f\left ( x \right ) \right )$ is

0%
$0$
0%
$1$
0%
$3$
0%
none of these

Explanation

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

$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$ and

$g\left( x \right) =\begin{cases} { x }^{ 2 }+1\qquad ;\qquad x\neq 0 \\ 4\qquad \qquad ;\qquad x=0 \end{cases}.$

Then

$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right)\right)}$ is

0%
$1$
0%
$4$
0%
$5$
0%
non-existent

Explanation

$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right) \right) } =g\left( f\left( 0 \right) \right)$

$=g(2)={ 2 }^{ 2 }+1=5$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\left ( 1-\tan \dfrac{x}{2} \right )\left ( 1-\sin x \right )}{\left ( 1+\tan \dfrac{x}{2} \right )\left ( \pi -2x \right )^{3}}$ is

0%
$0$
0%
$\displaystyle \frac{1}{32}$
0%
$\infty$
0%
$\displaystyle \frac{1}{8}$

Explanation

$\because \tan\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = \dfrac{1 - \tan\dfrac{x}{2}}{1+\tan\dfrac{x}{2}}$

$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( 1-\sin x \right )}{4\left ( \dfrac{\pi -2x}{4} \right )\left ( \pi -2x \right )^{2}}$

$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2} \right )\left(1-\cos \left ( \dfrac{\pi }{2}-x \right )\right)}{4.\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( \pi -2x \right )^{2}}$

$\therefore \displaystyle\lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2}\right )\cdot2\cdot\sin ^{2}\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )}{4\cdot\left (\dfrac{\pi }{4}-\dfrac{x}{2} \right )\cdot4^{2}\cdot\left ( \dfrac{\pi -2x}{4} \right )^{2}}$

$\displaystyle =\frac{1}{4}\times \frac{2}{16}=\frac{1}{32}$

Which one of the following statements is true?

0%
If $\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$ and $\displaystyle\lim_{x\rightarrow c}{f(x)}$ exist, then $\displaystyle\lim_{x\rightarrow c}{g(x)}$ exists.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$ exists, then $\displaystyle\lim_{x\rightarrow c}{f(x)}$ and $\displaystyle\lim_{x\rightarrow c}{g(x)}$ exist.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$ and $\displaystyle\lim_{x\rightarrow c}{f(x)}$ exist, then $\displaystyle\lim_{x\rightarrow c}{g(x)}$ also exists.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$ exists, then $\displaystyle\lim_{x\rightarrow c}{f(x)}$ and $\displaystyle\lim_{x\rightarrow c}{g(x)}$ also exist.

Explanation

For option A and B, take

$f(x)=x$ and

$g(x)=\dfrac { 1 }{ x }$ .
Limit of

$g(x)$ doesn't exist at

$x=0$ , but for

$\displaystyle \lim _{ x\to 0} f(x).g(x)$ and

$f(x)=x$ limit exists.

For option C,
Lets assume

$h(x) = f(x)+g(x)$ ,
Take limit on both sides as

$x\to c$

$\lim_{x\to c}h(x) = \lim_{x\to c}(f(x)+g(x))$
And it is given that

$\lim_{x\to c}f(x)$ exists.
Using sum law of limits:
If

$\lim_{x\to c}F(x)$ and

$\lim_{x\to c}G(x)$ exists, then

$\lim_{x\to c}(F(x)\pm G(x))$ also exists.

$\therefore \lim_{x\to c}(h(x) - f(x)) = \lim_{x\to c}(f(x) + g(x) - f(x)) = \lim_{x\to c}g(x)$ exists.

For option D, take

$f(x)=-\dfrac{cosx}{x^2}$ and

$g(x)=\dfrac { 1 }{ x^2 }$ .
Limit of

$g(x)$ and

$f(x)=-\dfrac{cosx}{x^2}$ doesn't exist at

$x=0$ , but for

$\displaystyle \lim _{ x\to 0} f(x)+g(x)$ limit exists.

Hence option C.

$\displaystyle \lim_{x\rightarrow 0}(f(x)\:g(x))$ exists for any functions

$f$ and

$g$ then

0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ and $\displaystyle \lim_{x\rightarrow a}g(x)$ exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ exist but $\displaystyle \lim_{x\rightarrow a}g(x)$ may not exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ may not exist but $\displaystyle \lim_{x\rightarrow a}g(x)$ exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ and $\displaystyle \lim_{x\rightarrow a}g(x)$ may not exist

Explanation

Take

$\displaystyle f(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$ and

$\displaystyle g(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$ . Then

$f(x)\:g(x)=1$ for

$x\neq 0$ . Hence

$\displaystyle \displaystyle \lim_{x\rightarrow 0}f(x)$ and

$\displaystyle \lim_{x\rightarrow 0}g(x)$ does not exist but

$\displaystyle\lim_{x\rightarrow 0}f(x)\:g(x)=1$
Hence, option 'D' is correct.

Evaluate

$\displaystyle \lim_{n\rightarrow \infty }\left [ \frac{n!}{n^{n}} \right ]^{1/n}$ .

0%
$\displaystyle\frac{1}{e}$
0%
$\displaystyle\frac{1}{t}$
0%
$\displaystyle\frac{1}{n}$
0%
$none\ of\ above$

Explanation

Let

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{n!}{n^{n}} \right )^{1/n}$

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{1.2.3.4...n}{n.nnn....n} \right )^{1/n}$

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \left ( \frac{1}{n} \right )\left ( \frac{2}{n} \right )\left ( \frac{3}{n} \right )...\left ( \frac{n}{n} \right ) \right )^{1/n}$

$\displaystyle\Rightarrow ln P=\lim_{n\rightarrow \infty }\frac{1}{n}\left ( \log \left ( \frac{1}{n} \right )+\log \left ( \frac{2}{n} \right )+...\log \left ( \frac{n}{n} \right ) \right )$

$\displaystyle P=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{n}\log \frac{r}{n}$

$=\int_{0}^{1}lnxdx=\left [ xlnx-x \right ]^{1}_{0}$

$=\displaystyle \left ( 0-1 \right )-\lim_{x\rightarrow 0 }\left ( xlnx \right )+0$

$=\displaystyle -1-\lim_{x\rightarrow 0 }\frac{lnx}{1/x}=-1-\lim_{x\rightarrow 0 }\frac{1/x}{\left ( -1/x^{2} \right )}$

$=\displaystyle -1-\lim_{x\rightarrow 0 }x=-1+0=-1\Rightarrow lnP=-1$

$\displaystyle P=e^{-1}=1/e$

$\displaystyle \lim_{x\rightarrow\infty}\left(\frac{\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt(1 - \cos x)+...\infty) - 1}}}{x^2}\right)$ equals to

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

Explanation

Let

$\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+...\infty)}}}=y$

$\Rightarrow y = \sqrt{(1 - \cos x) + y}$

$\Rightarrow y^2 - y + (\cos x - 1) = 0$

$\Rightarrow y = \displaystyle\frac{1 + \sqrt{5 - 4 \cos x}}{2}$

Now required limit is

$\lim_{x\rightarrow0}$

$\displaystyle\frac{1 + \sqrt{5 - 4 cosx - 1}}{2x^2}$

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

State whether the given statement is true or false

If f(x) satisfies the relation, f(x+y)=f(x)+f(y) for all

$x, y\epsilon R$ and f(1)=5, then find

$\sum_{n=1}^{m}f\left ( n \right )$ .

0%
True
0%
False

Explanation

We have,

$f(x+y)=f(x)+f(y)$
Put

$x=y=0$ we get,

$f(0)=0\Rightarrow$ (1)
Now,

$\displaystyle f'(x)= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)+f(h)-f(x)}{h}$

$\displaystyle f'(x)= \lim_{h\to 0}\frac{f(h)-f(0)}{h} =f'(0)$ , using (1)
Integrating we get,

$f(x) = f'(0)\cdot x+c$
Also

$f(0) = 0\Rightarrow c=0$
It is given that,

$f(1) =5\Rightarrow f'(0) = 5$
Hence

$f(x)=5x$

$\therefore \displaystyle \sum_{n=1}^m f(n) = 5\times \sum_{n=1}^{m}(n) =\frac{5}{2}m(m+1) =$ odd number

If [.] denotes, GIF , then

$\underset{x \rightarrow 0}{lt} \left( \left[\dfrac{2018 sin^{-1} x}{x}\right] + \left[\dfrac{2020x}{tan^{-1} x}\right]\right)$ =

0%
4038
0%
4037
0%
4036
0%
4039

The value of

$\displaystyle \lim_{x \rightarrow 0} \left(\dfrac{\sin x}{x}\right)^{1/x^{2}}$ is

0%
$e^{-1/6}$
0%
$e^{1/6}$
0%
$e^{-1/3}$
0%
$e^{1/3}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

$\tan ^{ 2 }{ x } > 1 \Rightarrow \tan { x } > \displaystyle\frac { 1 }{ \sqrt { 3 } }$ or

$\tan { x } < \displaystyle\frac { -1 }{ \sqrt { 3 } }$

$\Rightarrow \displaystyle\frac { -\pi }{ 2 } < x < \displaystyle\frac { -\pi }{ 6 }$ or

$\displaystyle\frac { \pi }{ 6 } < x < \displaystyle\frac { \pi }{ 2 }$

$\displaystyle f\left( x \right) =\begin{cases} \begin{matrix} 0, & -\dfrac { \pi }{ 2 } <x<-\dfrac { \pi }{ 6 } \end{matrix} \\ \begin{matrix} \dfrac { x }{ 2 } , & x=-\dfrac { \pi }{ 6 } \end{matrix} \\ \begin{matrix} x, & -\dfrac { \pi }{ 6 } <x<\dfrac { \pi }{ 6 } \end{matrix} \\ \begin{matrix} \dfrac { x }{ 2 } , & x=\dfrac { \pi }{ 6 } \end{matrix} \\ \begin{matrix} 0, & \dfrac { \pi }{ 6 } <x<\dfrac { \pi }{ 2 } \end{matrix} \end{cases}$

$\because$ obviously Assertion is incorrect and Reason is correct.

let a, b, c are non zero constant number then

$\lim_{r\rightarrow\infty}\displaystyle\frac{cos\displaystyle\frac{a}{r}-cos\displaystyle\frac{b}{r}cos\displaystyle\frac{c}{r}}{sin\displaystyle\frac{b}{r}sin\displaystyle\frac{c}{r}}$ equals to

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$\displaystyle\frac{a^2 + b^2 - c^2}{2bc}$
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$\displaystyle\frac{c^2 + a^2 - b^2}{2bc}$
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$\displaystyle\frac{b^2 + c^2 - a^2}{2bc}$
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independent of a, b and c

Explanation

$\displaystyle \lim _{ r\rightarrow \infty }{ \dfrac { \cos { \dfrac { a }{ r } -\cos { \dfrac { b }{ r } } \cos { \dfrac { c }{ r } } } }{ \sin { \dfrac { b }{ r } \sin { \dfrac { c }{ r } } } } }$

$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { \cos { \dfrac { a }{ r } -\dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b+c }{ r } \right) } +\cos { \left( \dfrac { b-c }{ r } \right) } \right) } }{ \dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b-c }{ r } \right) -\cos { \left( \dfrac { b+c }{ r } \right) } } \right) } }$

By Applying L Hospital,s Rule

$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { -a\sin { \dfrac { a }{ r } -\dfrac { 1 }{ 2 } \left( -\left( b+c \right) \sin { \left( \dfrac { b+c }{ r } \right) } -\left( b-c \right) \sin { \left( \dfrac { b-c }{ r } \right) } \right) } }{ \dfrac { -1 }{ 2 } { \left( b-c \right) }\sin { \left( \dfrac { b-c }{ r } \right) } +\dfrac { 1 }{ 2 } { \left( b+c \right) }\sin { \left( \dfrac { b+c }{ r } \right) } } }$

$(here\,\,\,\dfrac{-1}{r^2}$

$would\,\, cancel \,\,out\,\, from\,\, numerator \,\,and\,\, denominator)$

$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { { -a }^{ 2 }\cos { \dfrac { a }{ r } +\dfrac { 1 }{ 2 } \left( { \left( b+c \right) }^{ 2 }\cos { \left( \dfrac { b+c }{ r } \right) +{ \left( b-c \right) }^{ 2 }\cos { \left( \dfrac { b-c }{ r } \right) } } \right) } }{ \dfrac { -1 }{ 2 } { \left( b-c \right) }^{ 2 }\cos { \left( \dfrac { b-c }{ r } \right) } +\dfrac { 1 }{ 2 } { \left( b+c \right) }^{ 2 }\cos { \left( \dfrac { b+c }{ r } \right) } } }$

$\displaystyle =\dfrac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 2bc }$

Evaluate

$\displaystyle \lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n^{2}} \right )\left ( 1+\frac{2^{2}}{n^{2}} \right )\left ( 1+\frac{3^{2}}{n^{2}} \right )......\left ( 1+\frac{n^{2}}{n^{2}} \right ) \right ]^{1/n}$

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$2e ^\left(\dfrac{\pi - 4}{2}\right)$
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$2e ^\left(\dfrac{\pi - 2}{2}\right)$
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$2e^ \left(\dfrac{\pi - 4}{4}\right)$
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$2e^ \left(\dfrac{\pi - 4}{3}\right)$

Explanation

Let

$\displaystyle S=\frac { lim }{ n\rightarrow \infty } { \left[ \left( 1+\frac { 1 }{ { n }^{ 2 } } \right) \left( 1+\frac { { 2 }^{ 2 } }{ { n }^{ 2 } } \right) ...\left( 1+\frac { { n }^{ 2 } }{ { n }^{ 2 } } \right) \right] }^{ \frac { 1 }{ n } }$

$\displaystyle \Rightarrow logS=\frac { lim }{ n\rightarrow \infty } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ log } \left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } } \right)$

$\displaystyle =\int _{ 0 }^{ 1 }{ log } \left( 1+{ x }^{ 2 } \right) dx.=\left( xlog\left( 1+{ x }^{ 2 } \right) \right) _{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \frac { { 2x }^{ 2 } }{ 1+{ x }^{ 2 } } dx }$

$\displaystyle \Rightarrow log\left( \frac { S }{ 2 } \right) =2\left[ \int _{ 0 }^{ 1 }{ \frac { { dx } }{ 1+{ x }^{ 2 } } } -\int _{ 0 }^{ 1 }{ dx } \right] =2\left[ \frac { \pi }{ 4 } -1 \right] =\frac { \pi -4 }{ 2 }$

$\displaystyle \Rightarrow S=2{ e }^{ \left( \frac { \pi -4 }{ 2 } \right) }$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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