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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 9
If $$\overset { \pi }{ \underset { 0 }{\int } } x \, f(sin\,x)dx=A\overset { \pi /2 }{ \underset { 0 }{ \int } } f(sin\,x)dx,$$ then A is _____________.
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$$0$$
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$$\pi$$
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$$\pi/4$$
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$$2\pi$$
$$\displaystyle \lim _{ x\rightarrow \infty }{ \left[\dfrac{n}{n^{2}+1^{2}}+\dfrac{n}{n^{2}+2^{2}}+\dfrac{n}{n^{2}+3^{2}}+....+\dfrac{1}{n^{5}}\right] }$$
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$$\pi/4$$
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$$\tan^{-1}{(2)}$$
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$$\pi/2$$
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$$\tan^{-1}{(3)}$$
Find the derivative of $$\sqrt{\tan x}$$ with respect to $$x$$ using the first principle.
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$$\dfrac {sec^2x}{2\sqrt {tanx}}$$
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$$\dfrac {secx}{2\sqrt {tanx}}$$
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$$\dfrac {sec^2x}{\sqrt {tanx}}$$
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$$\dfrac {sec^2x}{2\sqrt {tan}}$$
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
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$$1$$
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$$-1$$
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$$\displaystyle \frac{1}{\sqrt{2}}$$
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$$\displaystyle \frac{1}{2}$$
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
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$$-1$$
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$$1/2$$
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$$-1/2$$
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$$1$$
The value of $$\displaystyle \lim_{\theta \rightarrow 0^{+}} \dfrac {\sin \sqrt {\theta}}{\sqrt {\sin \theta}}$$ is equal to
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$$0$$
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$$1$$
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$$-1$$
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$$4$$
$$f(x)=\sin x$$ and $$f^{\prime} (\pi)$$
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$$-1$$
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$$0$$
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$$1$$
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None of these
Explanation
$$f(x)=\sin x\\f^{\prime} (x)=\dfrac {d}{dx}\sin x\\f^{\prime} (x)=\cos x\\f^{\prime}(\pi)=\cos \pi=-1$$
If $$x+y=\sin (x-y)$$ then $$\dfrac { dy }{ dx } $$ is equal to
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$$\dfrac { 1 }{ 2 } $$
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$$0$$
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$$-1$$
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$$\dfrac { 1 }{ 3 } $$
Explanation
Given,
$$x+y=\sin (x-y)$$
$$x+y=\sin x\cos y-\cos x\sin y$$
$$1+\dfrac{dy}{dx}=\sin x(-\sin y)+\cos y(\cos x)-\cos x(\cos y)-(-\sin x)\sin y$$
$$1+\dfrac{dy}{dx}=-\sin x\sin y+\cos x\cos y+\sin x\sin y-\cos x\cos y$$
$$1+\dfrac{dy}{dx}=0$$
$$\therefore \dfrac{dy}{dx}=-1$$
If $$y = \sqrt{\dfrac{sec x-1}{sec x+1}}$$ then $$\dfrac{dy}{dx}$$ =
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$$\dfrac{1}{2} sec^2 \dfrac{x}{2}$$
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$$ sec^2 \dfrac{x}{2}$$
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$$\dfrac{1}{2} tan \dfrac{x}{2}$$
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$$tan \dfrac{x}{2}$$
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
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$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } f'\left( \dfrac { 1 }{ x } \right) =1$$
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$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } xf\left( \dfrac { 1 }{ x } \right) =2$$
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$$\underset { x\rightarrow { 0 }^{ + } }{ \lim } x^{ 2 }f'\left( x \right) =0$$
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$$\left| f\left( x \right) \right| \le 2$$ 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.
If $$ \mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) } $$ is non-
zero finite $$ ( \mathrm { a } > 0 ) , $$ then-
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L = 2 , b = 1 , c = 1
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$$
L = \frac { 1 } { 2 } , b = 1 , c = 1
$$
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L = 4 , b = - 1 , c = - 1
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$$
L = \frac { 1 } { 4 } , b = - 1 , c = - 1
$$
The solution the differential equation $$\cos x \sin y dx+ \sin x \cos y dy =0$$
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$$\dfrac{\sin x}{\sin y}=c$$
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$$\cos x+ \cos y=c$$
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$$\sin x + \sin y =c$$
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$$\sin x. \sin y=c$$
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 :
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0
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-1
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1
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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?
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$$2$$
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$$-2$$
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$$0$$
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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
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$$e^2$$
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$$e$$
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$$e^{-1}$$
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$$1$$
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.
$$\displaystyle\lim_{x\rightarrow a}\dfrac{\sqrt{x}+\sqrt{a}}{x+a}$$.
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$$-\dfrac{1}{\sqrt{a}}$$
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$$\dfrac{1}{{a}}$$
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$$\dfrac{1}{2\sqrt{a}}$$
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$$\dfrac{1}{\sqrt{a}}$$
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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{3x+1}{x+3}$$.
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$$\dfrac{1}{3}$$
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$$\dfrac{2}{3}$$
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$$\dfrac{5}{3}$$
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None of these
Explanation
Evaluate
$$\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}$$
as $$\to 0, Expression \to \dfrac {1}{3}$$, 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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2/3}-9}{x-27}$$.
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{5}$$
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None of these
Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{ax+b}{cx+d}, d\neq 0$$.
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$$\dfrac{a}{c}$$
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$$\dfrac{a}{d}$$
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$$\dfrac{b}{d}$$
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None of these
Explanation
$$\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}, d\neq 0$$
as $$x\to 0$$, 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.
$$\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}, x > 1$$.
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$$\dfrac{\sqrt{2}+1}{\sqrt{2}}$$
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$$\dfrac{\sqrt{2}-1}{\sqrt{2}}$$
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$$\dfrac{\sqrt{2}+1}{{2}}$$
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None of these
Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^2+x^2}-a}{x^2}$$.
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$$\dfrac{1}{\sqrt a}$$
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$$\dfrac{1}{\sqrt {2a}}$$
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$$\dfrac{1}{a}$$
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$$\dfrac{1}{2a}$$
Explanation
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^{2}+x^{2}}-a}{x^{2}}$$
As $$x\rightarrow 0$$, 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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{2x}{\sqrt{a+x}-\sqrt{a-x}}$$.
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$$-2\sqrt{a}$$
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$$\sqrt{a}$$
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$$2\sqrt{a}$$
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None of these
Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow a}\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$$.
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$$2\sqrt{a}$$
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$$2{a}$$
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$$2{a^{\frac 13}}$$
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None of these
Explanation
$$\displaystyle \lim_{x\rightarrow a}{\dfrac{x-a}{\sqrt x-\sqrt a}}$$
if $$x\rightarrow a$$, 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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}$$.
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$$\dfrac{1}{2\sqrt{a}}$$
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$$\dfrac{1}{2a\sqrt{a}}$$
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$$\dfrac{1}{2a}$$
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None of these
Explanation
$$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a^{2}+ ax}}$$
as $$x \rightarrow 0$$, 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.
$$\displaystyle\lim_{x\rightarrow 4}\dfrac{2-\sqrt{x}}{4-x}$$.
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{3}$$
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None of these
Explanation
$$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{4-\sqrt x}}$$
if $$x\rightarrow 4$$, 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.
$$\displaystyle\lim_{x\rightarrow 2}\dfrac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$$.
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{3}$$
0%
$$\dfrac{1}{4}$$
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$$\dfrac{1}{5}$$
Explanation
Evaluate $$\displaystyle \lim_{x\to 2}\dfrac {\sqrt {1+4x}-\sqrt {5+2x}}{(x-2)}$$
as $$x\to 2$$, 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.
$$\displaystyle\lim_{x\rightarrow 2}\dfrac{x-2}{\sqrt{x}-\sqrt{2}}$$.
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$$2\sqrt{3}$$
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$$2\sqrt{2}$$
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$$2\sqrt{5}$$
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None of these
Explanation
$$\displaystyle\lim_{x\rightarrow 2}\dfrac{(x-2)}{\sqrt{x}-\sqrt{2}}$$
as $$x\rightarrow 2$$, 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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{8^x-2^x}{x}$$.
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$$log 4$$
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$$log 6$$
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$$log 5$$
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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.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}$$.
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$$\dfrac{1}{\sqrt{2}}$$
0%
$$-\dfrac{1}{\sqrt{3}}$$
0%
$$-\dfrac{1}{{2}}$$
0%
$$-\dfrac{1}{\sqrt{2}}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}$$
As $$x\rightarrow 0$$, 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.
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$$2, -2$$.
0%
$$1, -1$$.
0%
$$1, 0$$.
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None of these
Explanation
Using formula $$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=na^{n-1}$$
$$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^9-a^9}{x-a}}=9.9^8=9$$
$$a^8=1\Rightarrow a=\pm 1$$
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