If $$y = \sqrt{\dfrac{sec x-1}{sec x+1}}$$ then $$\dfrac{dy}{dx}$$ =

$$\dfrac{1}{2} tan \dfrac{x}{2}$$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 9

$\overset { \pi }{ \underset { 0 }{\int } } x \, f(sin\,x)dx=A\overset { \pi /2 }{ \underset { 0 }{ \int } } f(sin\,x)dx,$ then A is _____________.

0%
$0$
0%
$\pi$
0%
$\pi/4$
0%
$2\pi$

$\displaystyle \lim _{ x\rightarrow \infty }{ \left[\dfrac{n}{n^{2}+1^{2}}+\dfrac{n}{n^{2}+2^{2}}+\dfrac{n}{n^{2}+3^{2}}+....+\dfrac{1}{n^{5}}\right] }$

0%
$\pi/4$
0%
$\tan^{-1}{(2)}$
0%
$\pi/2$
0%
$\tan^{-1}{(3)}$

Find the derivative of

$\sqrt{\tan x}$ with respect to

$x$ using the first principle.

0%
$\dfrac {sec^2x}{2\sqrt {tanx}}$
0%
$\dfrac {secx}{2\sqrt {tanx}}$
0%
$\dfrac {sec^2x}{\sqrt {tanx}}$
0%
$\dfrac {sec^2x}{2\sqrt {tan}}$

Explanation

$y=\sqrt{\tan x}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{\tan x}}\times \sec^2x$

$=\dfrac{\sec^2x}{2\sqrt{\tan x}}$

Hence, the answer is

$\dfrac{\sec^2x}{2\sqrt{\tan x}}.$

The value of

$\displaystyle n\xrightarrow { lim } \infty\frac{1.n+2.(n-1)+3.(n-2)+...+n.1}{{1}^{2}+{2}^{2}+...+{n}^{2}}$ is

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

Explanation

Now,

$1.n+2(n-1)+3.(n-2)+....+n.1$

The general term of the above series is

${ T }_{ r }=r(n-r+1)=r(n+1-r)$

Sum of series

$=S=\sum _{ r=1 }^{ n }{ { T }_{ r } } =\sum _{ r=1 }^{ n }{ r(n+1-r) }$

$S=\sum _{ r=1 }^{ n }{ nr } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\sum _{ r=1 }^{ n }{ r } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\cfrac { n(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 }$

Similarly,

${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ n }^{ 2 }=\cfrac { n(n+1)(2n+1) }{ 6 }$

$\lim _{ n\rightarrow \infty }{ \cfrac { 1.n+2(n-1)+3.(n-2)+...+n.1 }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+........{ n }^{ 2 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { { n }^{ 2 }(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 } }{ \cfrac { n(n+1)(2n+1) }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { n }{ 2 } +\cfrac { 1 }{ 2 } -\cfrac { 2n+1 }{ 6 } }{ \cfrac { 2n+1 }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ (\cfrac { 3n+3-2n-1 }{ 2n+1 } ) } =\lim _{ n\rightarrow \infty }{ (\cfrac { 3+\cfrac { 3 }{ n } -2-\cfrac { 1 }{ n } }{ 2+\cfrac { 1 }{ n } } ) } \\ =\cfrac { 3-2 }{ 2 } =\cfrac { 1 }{ 2 }$

The value of

$\displaystyle \lim _{ x\rightarrow 0 } \dfrac{1+\sin{x}-\cos{x}+\log{(1-x)}}{x^{3}}$ , is

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

The value of

$\displaystyle \lim_{\theta \rightarrow 0^{+}} \dfrac {\sin \sqrt {\theta}}{\sqrt {\sin \theta}}$ is equal to

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

$f(x)=\sin x$ and

$f^{\prime} (\pi)$

0%
$-1$
0%
$0$
0%
$1$
0%
None of these

Explanation

$f(x)=\sin x\\f^{\prime} (x)=\dfrac {d}{dx}\sin x\\f^{\prime} (x)=\cos x\\f^{\prime}(\pi)=\cos \pi=-1$

$x+y=\sin (x-y)$ then

$\dfrac { dy }{ dx }$ is equal to

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

Explanation

Given,

$x+y=\sin (x-y)$

$x+y=\sin x\cos y-\cos x\sin y$

$1+\dfrac{dy}{dx}=\sin x(-\sin y)+\cos y(\cos x)-\cos x(\cos y)-(-\sin x)\sin y$

$1+\dfrac{dy}{dx}=-\sin x\sin y+\cos x\cos y+\sin x\sin y-\cos x\cos y$

$1+\dfrac{dy}{dx}=0$

$\therefore \dfrac{dy}{dx}=-1$

$y = \sqrt{\dfrac{sec x-1}{sec x+1}}$ then

$\dfrac{dy}{dx}$ =

0%
$\dfrac{1}{2} sec^2 \dfrac{x}{2}$
0%
$sec^2 \dfrac{x}{2}$
0%
$\dfrac{1}{2} tan \dfrac{x}{2}$
0%
$tan \dfrac{x}{2}$

Explanation

Given,

$y=\sqrt{\dfrac{\sec x-1}{\sec x+1}}$

$=\sqrt{\dfrac{\sec x-1}{\sec x+1}} \times \sqrt{\dfrac{\sec x-1}{\sec x-1}}$

$=\sqrt{\dfrac {(sec x-1)^2}{(\sec x-1)(\sec x+1)}}$

$=\sqrt{\dfrac{(sec x-1)^2}{\sec ^2x-1}}$

$=\sqrt{\dfrac{(sec x-1)^2}{\tan ^2x}}$

$=\dfrac{sec x-1}{\tan x}$

$=\dfrac{sec x}{\tan x}-\dfrac{1}{\tan x}$

$=\csc x -\cot x$

$\therefore y=\csc x -\cot x$

Now,

$\dfrac{dy}{dx}=\dfrac{d}{dx}(\csc x -\cot x)$

$=-\cot \left(x\right)\csc \left(x\right)-\left(-\csc ^2\left(x\right)\right)$

$=-\cot \left(x\right)\csc \left(x\right)+\csc ^2\left(x\right)$

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

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

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

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

$=-\dfrac{1-\cos \left(x\right)}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}$

$=\dfrac{1}{\cos \left(x\right)+1}$

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

$=\dfrac{1}{2} sec^2 \dfrac{x}{2}$

Let

$f:(0, \infty)\to R$ be a differentiable function such that

$f'(x)=2-\dfrac{f(x)}{x}$ for all

$x\in (0, \infty)$ and

$f(1)\neq 1$ . Then

0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } f'\left( \dfrac { 1 }{ x } \right) =1$
0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } xf\left( \dfrac { 1 }{ x } \right) =2$
0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } x^{ 2 }f'\left( x \right) =0$
0%
$\left| f\left( x \right) \right| \le 2$ for all $X\in \left( 0,2 \right)$

Explanation

Here,

$f'(x)=2-\dfrac{f(x)}{x}$
or

$\dfrac{dy}{dx}+\dfrac{y}{x}=2$ [ i.e. linear differential equation in

$y$ ]
Integrating Factor,

$IF=e^{\displaystyle \int{\dfrac{1}{x}dx}}=e^{\log x}=x$

$\therefore$ Required solution is

$y.(IF)=\displaystyle \int{Q(IF)dx+C}$

$\Rightarrow y(x)=\int{2(x)dx+C}$

$\Rightarrow yx=x^2+C$

$\therefore y=x+\dfrac{C}{x}$

$[ \because C \neq 0$ , as

$f(1) \neq 1]$
(a)

$\displaystyle \lim_{x \rightarrow 0^+}{f' \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1-Cx^2)}=1$

$\therefore$ Option (a) is correct.
(b)

$\displaystyle \lim_{x \rightarrow 0^+}{xf \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1+Cx^2)}=1$

$\therefore$ Option (b) is correct.
(c)

$\displaystyle \lim_{x \rightarrow 0^+}{x^2f' (x)}=\displaystyle \lim_{x \rightarrow 0^+}{(x^2-C)}=-C\neq 0$

$\therefore$ Option (c) is correct.
(d)

$f(x)=x+\dfrac{C}{x}, C \neq 0$
For

$C > 0, \displaystyle \lim_{x\rightarrow 0^+}{f(x)=\infty}$

$\therefore$ Function is not bounded in

$(0, 2)$ .

$\therefore$ Option (d) is incorrect.

$\mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) }$ is non-
zero finite

$( \mathrm { a } > 0 ) ,$ then-

0%
L = 2 , b = 1 , c = 1
0%
$L = \frac { 1 } { 2 } , b = 1 , c = 1$
0%
L = 4 , b = - 1 , c = - 1
0%
$L = \frac { 1 } { 4 } , b = - 1 , c = - 1$

The solution the differential equation

$\cos x \sin y dx+ \sin x \cos y dy =0$

0%
$\dfrac{\sin x}{\sin y}=c$
0%
$\cos x+ \cos y=c$
0%
$\sin x + \sin y =c$
0%
$\sin x. \sin y=c$

Explanation

Given differential equation is

$\cos x \sin ydx+\sin x \cos ydy=0$

$\Rightarrow \cos x\sin y dx=-\sin x\cos y dy$

$\Rightarrow \dfrac{\cos x}{\sin x}dx=-\dfrac{\cos y}{\sin y}dy$
On integrating both sides, we get

$\int {\cot xdx}=\int{-\cot ydy}$

$\log \sin x=-\log \sin y+\log C$

$\Rightarrow \log \sin x \sin y=\log C$

$\Rightarrow \sin x. \sin y=C$

For

$x>y$ ,

$\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]}$ is :

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

If the function

$f(x)$ satisfies the relation

$f(x+y)=y\dfrac{|x-1|}{(x-1)}f(x)+f(y)$ with

$f(1)=2$ , then

$\displaystyle\lim_{x\rightarrow 1}f'(x)$ is?

0%
$2$
0%
$-2$
0%
$0$
0%
Limit do not exixst

Let

$f : R \to R$ be a differentiable function satisfying

$f'(3) + f'(2) = 0$ .
Then

$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$ is equal to

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

Explanation

$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$

$(1^{\infty}$ form

$)$

$\Rightarrow e^{\underset{x \to 0}{\lim} \dfrac{f(3+x)-f(2-x)-f(3)+f(2)}{x(1+f(2-x)-f(2))}}$

using L'Hopital

$\Rightarrow e^{\underset{x \to 0}{\lim}\dfrac{f'(3+x)+f'(2-x)}{-xf'(2-x)+(1+f(2-x)-f(2))}}$

$\Rightarrow e^{\dfrac{f'(3) + f'(2)}{1}} = 1$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow a}\dfrac{\sqrt{x}+\sqrt{a}}{x+a}$ .

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

Explanation

Evaluate

$\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}$

not an indeterminant Form so, limit can be found by simply putting unit of

$x$

$\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}=\dfrac {2\sqrt a}{2a}=\dfrac {1}{\sqrt a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{3x+1}{x+3}$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{5}{3}$
0%
None of these

Explanation

Evaluate

$\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}$

$\to 0, Expression \to \dfrac {1}{3}$ , not indeterminant form

so, Limit can be foord, by simple substituting value

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

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2/3}-9}{x-27}$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{5}$
0%
None of these

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{ax+b}{cx+d}, d\neq 0$ .

0%
$\dfrac{a}{c}$
0%
$\dfrac{a}{d}$
0%
$\dfrac{b}{d}$
0%
None of these

Explanation

$\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}, d\neq 0$

$x\to 0$ , Expression

$\to \dfrac {b}{d}$ , which is a determined Form, so Limit= value

$\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}=\dfrac {b}{a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}, x > 1$ .

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

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^2+x^2}-a}{x^2}$ .

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

Explanation

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^{2}+x^{2}}-a}{x^{2}}$

$x\rightarrow 0$ , it is

$\dfrac{0}{0}$ Form,

So, using rationalisation

$\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^{2}+x^{2}}-a)}{x^{2}}\times \dfrac{(\sqrt{a^{2}+x^{2}}+a)}{\sqrt{a^{2}+x^{2}}+a)}$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{(a^{2}+x^{2})-a^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}$

$((a+b)(a+b)=a^{2}-b^{2})$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{(\sqrt{a^{2}+x^{2}}+a)}=\dfrac {1}{\sqrt {a^2}+a}=\dfrac{1}{2a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{2x}{\sqrt{a+x}-\sqrt{a-x}}$ .

0%
$-2\sqrt{a}$
0%
$\sqrt{a}$
0%
$2\sqrt{a}$
0%
None of these

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow a}\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$ .

0%
$2\sqrt{a}$
0%
$2{a}$
0%
$2{a^{\frac 13}}$
0%
None of these

Explanation

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x-a}{\sqrt x-\sqrt a}}$

$x\rightarrow a$ , expression

$\rightarrow \dfrac{0}{0}$ , an indetermined form

using factorisation

$\displaystyle \lim_{x\rightarrow a}{\dfrac{(\sqrt{x}-\sqrt a)(\sqrt x+\sqrt a)}{(\sqrt x-\sqrt a)}}=2\sqrt a$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}$ .

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

Explanation

$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a^{2}+ ax}}$

$x \rightarrow 0$ , it is

$\dfrac{0}{0}$ form

So, using rationalization

$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}- \sqrt{a}}{x \sqrt{a^{2}+ax}} \dfrac{(\sqrt{a+x}+ \sqrt{a})}{(\sqrt{a+x}+ \sqrt{a})}$

$\displaystyle \lim_{x \rightarrow 0} \dfrac{x}{x \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}$

$\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{ \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}$

$= \dfrac{1}{2a \sqrt{a}}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 4}\dfrac{2-\sqrt{x}}{4-x}$ .

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
None of these

Explanation

$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{4-\sqrt x}}$

$x\rightarrow 4$ , expression

$\rightarrow \dfrac{0}{0}$ , an indetermined form

using factorisation

$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{2^2-(\sqrt x)^2}}=\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{(2-\sqrt x)(2+\sqrt x)}}=\dfrac{1}{4}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 2}\dfrac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$ .

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$

Explanation

Evaluate

$\displaystyle \lim_{x\to 2}\dfrac {\sqrt {1+4x}-\sqrt {5+2x}}{(x-2)}$

$x\to 2$ , it is

$\dfrac {0}{0}$ From so,

using rationalisation

$\displaystyle \lim_{x\to 2}\dfrac {(\sqrt {1+4x}-\sqrt {5+2x}) (1+4x)+\sqrt {5+2x}}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$

$\displaystyle \lim_{x\to 2}\dfrac {((1+4x)-(5+2x))}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$

$\displaystyle \lim_{x\to 2}\dfrac {2(x-2)}{(x-2)(6)}=\dfrac {1}{3}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 2}\dfrac{x-2}{\sqrt{x}-\sqrt{2}}$ .

0%
$2\sqrt{3}$
0%
$2\sqrt{2}$
0%
$2\sqrt{5}$
0%
None of these

Explanation

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

$x\rightarrow 2$ , it is

$\dfrac{0}{0}$ Form,

So, using Factorisation

$\displaystyle\lim_{x\rightarrow 2}\dfrac{((\sqrt x)^2-(\sqrt 2)^2)}{\sqrt{x}-\sqrt{2}}=\displaystyle\lim_{x\rightarrow 2}\dfrac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})}$

$\displaystyle \lim_{x\to 2}\sqrt x +\sqrt 2=2\sqrt{2}$

Evaluate the following limit.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{8^x-2^x}{x}$ .

0%
$log 4$
0%
$log 6$
0%
$log 5$
0%
None of these

Explanation

we know

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{a^x-1}{x}}=\log a$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-2^x}{x}}=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1+1-2^x}{x}}$

$=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1}{x}}-\displaystyle \lim_{x\rightarrow 0}{\dfrac{2^x-1}{x}}=\log 8-\log 2$

$=\log 8/2=\log 4$ .

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}$ .

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

Explanation

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}$

$x\rightarrow 0$ , it is

$\dfrac{0}{0}$ from, so,

using rationalisation

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}\dfrac{(\sqrt{2-x}+\sqrt{2+x})}{(\sqrt{2-x}+\sqrt{2+x})}$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x})}}$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{-2x}{x(\sqrt{2-x}+\sqrt{2+x})}}=\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{\sqrt2}$

Evaluate the following limits.
If

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^9-a^9}{x-a}=9$ , find all possible values of a.

0%
$2, -2$ .
0%
$1, -1$ .
0%
$1, 0$ .
0%
None of these

Explanation

Using formula

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=na^{n-1}$

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^9-a^9}{x-a}}=9.9^8=9$

$a^8=1\Rightarrow a=\pm 1$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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