If $$y=(x+\sqrt{x^{2}+a^{2}})^{n}$$ then $$\dfrac{dy}{dx}=$$

$$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$$

$$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$$

If $$(cos x)^y = (sin y)^x$$, then $$\dfrac{dy}{dx}$$ =

$$\dfrac{log(siny) + y tan x}{ log (cos x) - x cot y}$$

$$\dfrac{log(siny) - y tan x}{ log (cos x) + cot y}$$

$$\dfrac{log(siny) }{ log (cos x) }$$

$$\dfrac{log(cos x) }{ log (sin y) }$$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 8

Find the value of limit

$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 } }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 } } = }$ .

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

Explanation

$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +2\sin { x } -\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -2\sin { x } -\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin { x } \left( \sin { x } +1 \right) -1\left( \sin { x } +1 \right) }{ 2\sin { x } \left( \sin { x } -1 \right) -1\left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \left( 2\sin { x } -1 \right) \left( \sin { x } +1 \right) }{ \left( 2\sin { x } -1 \right) \left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \sin { x } +1 }{ \sin { x } -1 } } \\ =\frac { \frac { 1 }{ 2 } +1 }{ \frac { 1 }{ 2 } -1 } \\ =-3$

$f(x+y)=2f(x).f(y)$ for all

$x,y$ , where

$f'(0)=3$ and

$f'(4)=2$ then

$f'(4)=3$ is equal to

0%
6
0%
12
0%
4
0%
3

$y=(x+\sqrt{x^{2}+a^{2}})^{n}$ then

$\dfrac{dy}{dx}=$

0%
$y$
0%
$ny$
0%
$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$
0%
$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$

Explanation

$y={ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n }\quad$

$\cfrac { dy }{ dx } =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\cfrac { d }{ dx } \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right)$

$=n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\left( 1+\cfrac { x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right) =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }\left( \cfrac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } +x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right)$

$=\cfrac { n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n } }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\cfrac { n.y }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } }$

$f(x)$ is the integral of

$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$ . Find

$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right) }$ , where

$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$

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

Explanation

$f(x)=\dfrac{2\sin x-\sin 2x}{x^3}$

$x\neq 0$

$f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}$

$=\dfrac{2\sin x}{x}\left(\dfrac{1-\cos x}{x^2}\right)$

$\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)$

$\Rightarrow \displaystyle 4\cdot 1 \lim_{x\rightarrow 0}{\left(\dfrac{\sin^2(x/2)}{4\times (x/2)^2}\right)}$

$=1$

$\lim\limits_{x\rightarrow0}\dfrac{x\tan 2x-2x\tan x}{\left(1-\cos 2x\right)^{2}}=$

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

Explanation

$\underset{x\rightarrow 0}{\lim}\dfrac{x\tan2x-2x\tan x}{(1-\cos 2x)^2}$

$\underset{x\rightarrow 0}{\lim}\dfrac{x[\tan2x-2\tan x]}{(1-\cos 2x)^2}$

$\underset{x\rightarrow 0}{\lim}x\dfrac { \left[ \frac { 2\tan { x } }{ 1-{ \tan { x } }^{ 2 } } -2\tan { x } \right] }{ { \left( 2{ \sin { x } }^{ 2 } \right) }^{ 2 } }$

$\underset{x\rightarrow 0}{\lim}{ \dfrac { 2x\tan { x\left[ \dfrac { 1 }{ 1-{ \tan { x } }^{ 2 } } -1 \right] } }{ 4{ \sin { x } }^{ 4 } } }$

$\underset{x\rightarrow 0}{\lim}\dfrac { \dfrac { x\tan { x } }{ \left( 1-{ \tan { x } }^{ 2 } \right) } \left[ 1-1+{ \tan { x } }^{ 2 } \right] }{ 2{ \sin { x } }^{ 4 } }$

$\underset{x\rightarrow 0}{\lim} {\dfrac { x{ \tan^3 { x } }^{ 3 } }{ ( 1-{ \tan^2 { x } )}.2{ \sin^4 { x } } } }$

Divide num and den by

$x^4$

$\underset{x\rightarrow 0}{\lim}\dfrac{\dfrac{\sin^3 x}{\cos^3 x}.\dfrac{x}{x^4}}{2\dfrac{\sin^4x}{x^4}(1-\tan^2 x)}$

$\underset{x\rightarrow 0}{\lim}{ \dfrac { { \left( \dfrac { \sin { x } }{ x } \right) }^{ 3 }\dfrac { 1 }{ { \cos ^{ 3 }{ x } } } }{ 2{ \left( \dfrac { \sin { x } }{ x } \right) }^{ 4 }\left( 1-{ \tan^2 { x } } \right) } }$

Apply limit

${ \dfrac { { \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 3 }.\dfrac { 1 }{ { \cos ^{ 3 }{1 } } } }{ 2{ \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 4 }\left( 1-{ \tan^2 { 1 } }\right) } }=\dfrac{1.1}{2.1(1-0)}$

$=\dfrac{1.1}{2.1(1-0)}$

$=\dfrac{1}{2}$

$\therefore \underset{x\rightarrow 0}{\lim}\dfrac{x \tan 2x- 2x\tan x}{(1-\cos 2x)^2}=\dfrac{1}{2}$

$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}=$ __________.

0%
$\cos\sqrt{x}$
0%
$\dfrac{1}{2\sin \sqrt{x}}$
0%
$\dfrac{\cos \sqrt{x}}{2\sqrt{x}}$
0%
$\sin\sqrt{x}$

Explanation

$Lt_{x\rightarrow 0}\cfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{(x+h)-x}$

Let

$f(x)=\sin\sqrt{x}$

$Lt_{h\rightarrow 0}\cfrac{\sin{x+h}-\sin\sqrt{x}}{(x+h)-x}=Lt_{x\rightarrow 0}f\cfrac{(x+h)-f(x)}{h}=f^1(x)\\ \Rightarrow \cos\sqrt x\times\cfrac{1}{2\sqrt x}\\ \Rightarrow \cfrac{\cos\sqrt x}{2\sqrt x}$

$2f(\sin x)+\sqrt {2}f(\cos x)=\tan x,\ (x> 0)$ , then

$\displaystyle \lim _{ x\rightarrow 1 }{ \sqrt { 1-x } f\left( x \right) = }$

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

$\dfrac{d}{dx}\left[\dfrac{tan x - cot x}{tan x + cot x}\right]=$

0%
2 sin2x
0%
-2 sin2x
0%
2 cos 2x
0%
-2cos 2x

Explanation

$\dfrac{d}{dx}\left(\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}}{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}}\right)=\dfrac{d}{dx}\left(\dfrac{\sin^2x-\cos^2x}{\sin^2x+\cos^2x}\right)$

$=\dfrac{d}{dx}(\sin^2x-\cos^2x)=2\sin x\cos x+2\cos x\sin x=2\sin 2x\Rightarrow (A)$

$\underset { x\rightarrow { 0 } }{ lim } \dfrac { \left( ax+b \right) -\sqrt { 4+\sin x } }{ \tan\quad x } =\dfrac { 27 }{ 4 } ~where ~a,b\in R$ then the value of

0%
$a = 2~ and~ b = 7$
0%
$a = -2~and~ b = 7$
0%
$a = 7~ and~ b = 2$
0%
$a=7~ and ~b = -2$

Evaluate:

$\underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } }$

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

Explanation

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

$=\underset{x\rightarrow 0}{lt}\dfrac{x\dfrac{2\tan x}{1-\tan^2x}-2x\tan x}{(1-1+2\sin^2x)^2}$ (since

$\tan 2\theta =\dfrac{2\tan \theta}{1-\tan^2\theta}$ )

$=\underset{x\rightarrow 0}{lt}\dfrac{2x\tan x(1-1+\tan^2x)}{(1-\tan^2x)y\sin^4x}=\underset{x\rightarrow 0}{lt}\dfrac{x\tan^3x}{2\sin^4x}\underset{x\rightarrow 0}{lt}\dfrac{1}{1-\tan^2x}$

$=\underset{x\rightarrow 0}{lt}\dfrac{\sin^3x/\cos^3x}{2\sin^3x(\sin x/x)}\times 1=\underset{x\rightarrow 0}{lt}\dfrac{1}{2\cos^3x}\underset{x\rightarrow 0}{lt}\dfrac{1}{\sin x/x}=\dfrac{1}{2}\times 1$

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

$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$ equals:

0%
$\frac { 1 } { 24 }$
0%
$\frac { 1 } { 16 }$
0%
0
0%
$\frac { 1 } { 4 }$

The solution of the differential equation

$\left( \dfrac { d y } { d x } \right) ^ { 2 } - 3 x \left( \dfrac { d y } { d x } \right) - 2 y = 8$ is

0%
$y = 2 x ^ { 2 } + 4$
0%
$y = 2 x ^ { 2 } - 4$
0%
$y = 2 x + 4$
0%
$y = 2 x - 4$

Explanation

By option Verification

Put

$\dfrac{dy}{dx}=4x$

substitute this in the equation

$\implies y=2x^2-4$

$\underset {x\rightarrow0}{ lim } \dfrac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^2}$ equals

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

Let

$f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}$ . If the function

$f(x)$ be continous at

$x=0$ , then

$k=$

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

Explanation

We have,

${{x}^{2}}+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when,\,x\ge 0\,$

$-{{x}^{2}}-k\,\,\,\,\,\,\,\,\,\,\,\,\,when,x\le 0$

Given that,

$L.H.L=R.H.L=f\left( x \right)$

At point $x=0$

${{x}^{2}}+k=-{{x}^{2}}-k=0$

$\Rightarrow {{x}^{2}}+k=0$

$\Rightarrow k=-{{x}^{2}}$

$\Rightarrow k=0$

Hence, this is the answer.

The value of

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

$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$ , is

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

Explanation

$\mathop {f\left( x \right)}\limits_{\lim \,\,x \to 0} = \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}$

$= \frac{{ - \sin \left( {\sin x} \right)\cos x - \sin x}}{{4{x^3}}}$

$= \frac{{ - \left( {\sin x} \right){{\cos }^2}x + \sin \left( {\sin x} \right)\sin x - \cos x}}{{12{x^2}}}$

$= \frac{{\sin \left( {\sin x} \right){{\cos }^3}x + \cos \left( {\sin x} \right)\cos 2x + \cos \left( {\sin x} \right)\sin x + \sin \left( {\sin x} \right)\cos x + \sin x}}{{24x}}$

$= \frac{{1 + 1 + Q + 1 + 1}}{{24}} = \frac{1}{6}$

The value of

$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } } \left( a>b \right)$ is

0%
$\dfrac {1}{4a}$
0%
$\dfrac {1}{a\sqrt {a-b}}$
0%
$\dfrac {1}{2a\sqrt {a-b}}$
0%
$\dfrac {1}{4a\sqrt {a-b}}$

Explanation

We have,

$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } }$

This is the

$\dfrac{0}{0}$ form.

Apply L-hospital rule,

$\lim_{x\to a}\dfrac{\dfrac{1}{2\sqrt {x-b}}-0}{2x-0}$

$\lim_{x\to a}\dfrac{1}{4x\sqrt {x-b}}$

$=\dfrac{1}{4a\sqrt {a-b}}$

Hence, this is the answer.

$\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }$ equals :

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

Explanation

$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 27+x \right) }^{ 1/3 }-3 }{ 9-{ \left( 27+x \right) }^{ 2/3 } }$

$=\underset { x\rightarrow 0 }{ lt } \dfrac { \dfrac { 1 }{ 3 } { \left( 27+x \right) }^{ -2/3 } }{ \dfrac { -2 }{ 3 } { \left( 27+x \right) }^{ -1/3 } }$

$=\underset { x\rightarrow 0 }{ lt } \dfrac { -1 }{ 2 } \dfrac { { \left( 27+x \right) }^{ 1/3 } }{ { \left( 27+x \right) }^{ 2/3 } }$

$=\dfrac { -1 }{ 2 } \left( \dfrac { 3 }{ 9 } \right) ,\quad \dfrac { -1 }{ 6 }$ [A]

A curve in the

$1^{st}$ quadrant passes through

$(1,1)$ . Its drifferential equation is

$(y-xy^{2})dx+(x+x^{2}y^{2})dy=0$ . Hence the equation of the curve is

0%
$y-\dfrac{1}{xy}=\ln y$
0%
$y-\dfrac{1}{xy}=\ln x$
0%
$y-xy=\ln y$
0%
$y-xy=\ln x$

$(cos x)^y = (sin y)^x$ , then

$\dfrac{dy}{dx}$ =

0%
$\dfrac{log(siny) + y tan x}{ log (cos x) - x cot y}$
0%
$\dfrac{log(siny) - y tan x}{ log (cos x) + cot y}$
0%
$\dfrac{log(siny) }{ log (cos x) }$
0%
$\dfrac{log(cos x) }{ log (sin y) }$

$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } } \right] }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right) }{ { x }^{ 3 }-{ x }^{ -1 } } \right] =$

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

Explanation

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4}{x^2-x^{-1}}-\dfrac{1-3x+x^2}{1-x^3}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4x}{x^3-1}+\dfrac{1-3x+x^2}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{x^2+x+1}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)x}{x^4-1}\right]$

$\displaystyle\lim_{x\rightarrow 0}\left[\dfrac{(x-1)(x^2+x+1)}{x^2+x+1}+3x\right]$ as

$(x^3-1)2(x-1)(x^2+x+1)$

$\displaystyle\lim_{x\rightarrow 1}[(x-1)+3x]=(1-1)+3(1)=3$

Answer

$=3$ .

1182108_1301374_ans_e0d4cf58fd6f414f8639427665c36517.jpg

Let

$U_{ { n } }=\dfrac { n! }{ (n+2)! }$ where

$n \in N .$ If

$S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }$ then

$\lim _ { n \rightarrow \infty } \mathrm { S } _ { n }$ equals :

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

Explanation

$\begin{array}{l} { U_{ n } }=\frac { { n! } }{ { \left( { n+2 } \right) ! } } =\frac { 1 }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } =\frac { { \left( { n+1 } \right) -\left( { n+1 } \right) } }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } \\ =\frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } \\ { S_{ n } }=\sum _{ n=1 }^{ n }{ { U_{ n } } } =\left[ { \left( { \frac { 1 }{ 2 } -\frac { 1 }{ 3 } } \right) +\left( { \frac { 1 }{ 3 } -\frac { 1 }{ 4 } } \right) +............+\left( { \frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } } \right) } \right] \\ { S_{ n } }=\frac { 1 }{ 2 } -\frac { 1 }{ { n+2 } } \\ { \lim }_{ n\to \infty }{ S_{ n } }=\frac { 1 }{ 2 } \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

$\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x }$ is :

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

Explanation

Let

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

$=\displaystyle \lim_{x\rightarrow 0} \frac { { 4 sin^4 x}}{\frac{1 - \tan^2 x}{2 x \tan x - 2x \tan^3 x - 2x \tan x}}$

$=\displaystyle \lim_{x\rightarrow 0} \frac {4 sin ^4 x ( 1 - \tan^2 x)}{- 2 x \tan^3 x}$

$=\displaystyle \lim_{x\rightarrow 0} \frac {2 sin^4 x}{-x \frac {sin^3 x}{cos x}}( 1 - \tan^2 x)$

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

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

$L\displaystyle = -2(1)(1)$

$L\displaystyle = -2$

The differential of

$f(x)=\sqrt{\dfrac{2-x}{2+x}}$ at

$x=0$ and

$\delta x=0.15$ is

0%
$0.07$
0%
$0.075$
0%
$-0.075$
0%
$0.15$

The value of

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

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

Explanation

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right) \cdot \sin 5x}}{{{x^2}\sin 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{ 2{{\sin }^2}x \cdot \sin 5x}}{{{x^2}\sin 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin x}}{x}} \right)}^2} \cdot \left( {\dfrac{{\sin 5x}}{{5x}}} \right) \cdot 5x}}{{\left( {\dfrac{{\sin 3x}}{{3x}}} \right) \cdot 3x}}$

$= \dfrac{{10}}{3}$ .

Let

$f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2$ and

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

$f(-2)=$

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

Explanation

$\underset{x\rightarrow 0}{lim} f(x)=2$

$\Rightarrow \underset{x\rightarrow 0}{lim} \dfrac{ax+b}{x+1}=2$

$\Rightarrow \dfrac{b}{1}=2$

$\therefore b=2$

Its is also given that

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

$\Rightarrow \underset{x\rightarrow \infty}{lim} \dfrac{ax+b}{x+1}=1$

$\Rightarrow \underset{x\rightarrow \infty}{lim}\dfrac{a+\dfrac{b}{x}}{1+\dfrac{1}{x}}=1$

$\Rightarrow \dfrac{a+0}{1+0}=1$

$\therefore a=1$

Substituting the values of a and b in

$f(x)=\dfrac{ax+b}{x+1}$ , we get,

$f(x)=\dfrac{x+2}{x+1}$

$\therefore f(-2)=\dfrac{-2+2}{-2+1}=0$

Evaluate the limit,

$\mathop {\lim }\limits_{x \to 0} \frac{{x({{(1 + x)}^{1/x}} - e)}}{{x({{(1 + {x^2})}^{1/{x^2}}} - e)}}$

0%
0
0%
1
0%
2
0%
DNE

Evaluate

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

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

Explanation

We have,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

We know that

$\cos 2x=1-2\sin^2x$

Therefore,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - (1-2\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (2\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - (1-2\sin^2 (\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{2\sin^2 (\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{2\sin^2 (\sin^2 x)}{\sin^4 x}\times \dfrac{\sin^4 x}{x^4}$

$2\mathop {\lim }\limits_{x \to 0} \left(\cfrac{\sin^2 (\sin^2 x)}{(\sin^2 x)}\right)^2\times \dfrac{\sin^4 x}{x^4}$

We know that

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

Therefore,

$\Rightarrow 2\times 1\times 1$

$\Rightarrow 2$

Hence, this is the answer.

$\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } - x + 1 } - a x - b \right) = 0,$ then the values of

$a$ and

$b$ are given by

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

Explanation

We have,

$\begin{array}{l} { \lim }_{ x\to \infty }\left( { \sqrt { { x^{ 2 } }-x+1 } -ax-b } \right) =0 \\ Put\, \, x=\cfrac { 1 }{ t } \\ { \lim }_{ t\to { 0^{ + } } }\left( { \sqrt { \cfrac { 1 }{ { { t^{ 2 } } } } -\cfrac { 1 }{ t } +1 } -\cfrac { a }{ t } -6 } \right) =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { { { \left( { 1-t+{ t^{ 2 } } } \right) }^{ \cfrac { 1 }{ 2 } } }-a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-\cfrac { t }{ 2 } +\cfrac { { { t^{ 2 } } } }{ 2 } } \right) -a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-a } \right) +t\left( { -b-\cfrac { 1 }{ 2 } } \right) } }{ t } =0 \\ 1=a \\ and\, \, b=-\cfrac { 1 }{ 2 } \\ Option\, \, \, C\, \, \, is\, \, correct\, \, answer. \end{array}$

$\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan x }{ \cos (x+\pi /4) }$ is

0%
$4$
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$8 \sqrt { 2 }$
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$8$
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$4 \sqrt { 2 }$

Let

$f(x)=\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{r=0}\dfrac{x}{(rx+1)\{(r+1)x+1\}}$ , then?

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$f(x)$ is continuous but not differentiable at $x=0$
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$f(x)$ is both continuous and differentiable at $x=0$
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$f(x)$ is neither continuous nor differentiable at $x=0$
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$f(x)$ is a periodic function

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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