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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 2 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 2
Find $$\dfrac{{dy}}{{dx}}$$ of the following $$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$$
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$$x+2x^2+6x^3+(n-1)(n-2)x^{n-3}$$
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$$2+6x+(n-1)x^{n-2}$$
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$$2+6x+(n-1)(n-2)x^{n-3}$$
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None
Explanation
$$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$$
$$\dfrac{{dy}}{{dx}} = 2 + 6x + \left( {n - 1} \right)\left( {n - 2} \right){x^{n - 3}}$$
If $$f(x) = sin x$$, find $$\frac{dy}{dx}$$.
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$$cosx$$
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$$-cosx$$
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$$cotx$$
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$$-cot^2x$$
Explanation
$$y=\sin x\Rightarrow \dfrac{dy}{dx}=\cos x$$
$$x^{\frac{1}{2}} + 1= t$$
differentiate w.r.t. x
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$$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$$
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$$\dfrac{dt}{dx} = \dfrac{1}{4\sqrt{x}}$$
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$$\dfrac{dt}{dx} = \dfrac{1}{8\sqrt{x}}$$
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$$\dfrac{dt}{dx} = \dfrac{1}{16\sqrt{x}}$$
Explanation
$$x^{\frac{1}{2}} + 1= t$$
$$\dfrac{dt}{dx} = \dfrac{1}{2}(x) ^{\frac{1}{2}-1} + 0$$
$$\frac{1}{2}(x) ^{\frac{-1}{2}}$$
$$\dfrac{1}{2x^{\frac{1}{2}}}$$
$$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$$
Differentiate: $$x^{100} + \sin x - 1$$
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$$100x^{99}-\cos x$$
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$$100x^{99}+\cos x$$
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$$x^{99}+\cos x$$
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$$100x^{99}+\sin x$$
Explanation
$$\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} +\sin x -1 \right )$$
$$=\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} \right )+\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( \sin x \right )-\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( 1 \right )$$
$$=100x^{100-1}+\cos x -0$$
$$=100x^{99}+\cos x$$
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$ equals
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$$\dfrac{7}{3}$$
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$$\dfrac{10}{3}$$
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$$\dfrac{14}{3}$$
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$$\dfrac{1}{3}$$
Explanation
We have,
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7x}{\dfrac{\sin3x}{3x}\times 3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7}{\dfrac{\sin3x}{3x}\times 3}$$
$$=\dfrac{1\times 7}{1\times 3}$$
$$=\dfrac{7}{3}$$
Hence, this is the answer.
Consider the differential equation $$\frac { d y } { d x } = \cos x$$
Then we observe that
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$$y = \sin x$$
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$$y = \sin x + 2$$
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$$y = \sin x - \frac { 1 } { 2 }$$
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$$y = \sin x + c$$
Explanation
$$dy = \cos x\>dx\\\int dy=\int \cos x\>dx\\y=\sin x+C$$
We must add constant "C" as its family of curves.
Which of the following statement is not correct
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } } $$
Explanation
All the result other than D are correct.
The last result will hold only when $$g(x)\ne 0$$ and $$\lim\limits_{x\to c} g(x)\ne 0$$ for $$x$$ belonging to a neighbourhood of $$c$$.
A, B, C are general results which are always true.
Evaluate the following limit :
$$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$$
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$$0$$
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$$1$$
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$$2$$
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none of these
Explanation
$$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$$
$$=lim_{x\rightarrow 0} \dfrac{2\sin^2 x}{x^2}$$
$$=2lim_{x \rightarrow 0} (\dfrac{\sin x}{x} \times \dfrac{\sin x}{x})$$
$$=2lim_{x\rightarrow 0} \dfrac{\sin x}{x} \times lim_{x\rightarrow 0} \dfrac{\sin x}{x}=2(1)(1)=2$$
Evaluate
$$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$
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$$\cfrac { 1 }{ 2 } $$
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$$1$$
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$${ 3 }^{ 2 }$$
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$$\cfrac { 1 }{ { 2 }^{ 2 } } $$
Explanation
$$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$
$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}$$
$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{2}(1+\dfrac{1}{n})=\dfrac{1}{2}$$
if $$y=5x^2+8x$$ find $$\dfrac {dy}{dx}$$
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$$10x+8$$
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$$5x+8$$
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$$10x^2+8x$$
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none of these
Explanation
The Equation is $$y=5x^2+8x$$
$$\dfrac {dy}{dx}=\dfrac d{dx}(5x^2+8x)\\\dfrac{dy}{dx}=10x+8$$
If $$y=e^{\sin \sqrt x}$$ then $$\dfrac{dy}{dx}=?$$
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$$e^{\sin \sqrt x}.\cos \sqrt x$$
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$$\dfrac{e^{\sin \sqrt x}\cos \sqrt x}{2\sqrt x}$$
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$$\dfrac{e^{\sin \sqrt x}}{2\sqrt x}$$
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$$none\ of\ these$$
If $$y=x^2\sin \dfrac{1}{x}$$ then $$\dfrac{dy}{dx}=?$$
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$$x\sin \dfrac{1}{x}-\cos \dfrac{1}{x}$$
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$$-\cos \dfrac{1}{x}+2x\sin \dfrac{1}{x}$$
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$$-x\sin \dfrac{1}{x}+\cos \dfrac{1}{x}$$
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$$none\ of\ these$$
If $$(x+y)=\sin (x+y)$$ then $$\dfrac{dy}{dx}=?$$
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$$-1$$
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$$1$$
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$$\dfrac{1-\cos (x+y)}{\cos^2(x+y)}$$
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$$none\ of\ these$$
If $$x=a\cos^{2}\theta, y=b\sin^{2}\theta$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{-a}{b}$$
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$$\dfrac{a}{b}\cot \theta$$
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$$\dfrac{-b}{a}$$
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$$none\ of\ these$$
If $$y=\cos^2 x^3$$ then $$\dfrac{dy}{dx}=?$$
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$$-3x^2\sin (2x^3)$$
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$$-3x^2\sin^2x^3$$
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$$-3x^2\cos^2(2x^3)$$
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$$none\ of\ these$$
lf $$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$ is finite then the value of $$a,b$$ respectively are
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$$5\, -4$$
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$$-5,\ -4$$
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$$-4,\ 3$$
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$$4,\ 5$$
Explanation
$$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$
As $${ x\rightarrow 0 }$$, denominator tends to 0, so the numerator also tends to 0.
$$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$$
at $$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$$
$$\Rightarrow 1+a+b=0$$
$$\Rightarrow a+b=-1$$
Now, put $$a = -4 $$
$$ \Rightarrow b = 3$$
Option C satisfies above equation.
$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$$
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$$\pi$$
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$$ 2\pi$$
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$$\displaystyle \frac{\pi}{2}$$
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$$\displaystyle \frac{2}{\pi}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$$
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) } $$
It is of the form $$\displaystyle \dfrac{0}{0}$$, so applying L-Hospital's rule
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } } $$
$$=\displaystyle \dfrac{2}{\pi}$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}$$=
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{3}{2}$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{1}{4}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}$$
$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$$...... As$$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$$
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$$1$$
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$$0$$
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$$-1$$
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$$\displaystyle \dfrac{1}{2}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$$
$$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$$
$$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) } $$
$$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$$
Solve :
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{4}{3}$$
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$$0$$
Explanation
Solve : $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$
$$=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)$$
$$=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}$$
$$=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}$$
$$=\dfrac34$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=$$
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$$\dfrac{10}{3}$$
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$$\dfrac{3}{10}$$
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$$\dfrac{6}{5}$$
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$$\dfrac{5}{6}$$
Explanation
Using, $$1 - cos 2x = 2{sin}^{2} x $$,
The expression transforms to,
$$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x} $$
Rewriting the expression in a different form,
$$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3} $$
Therefore, the limit to the expression is $$ \dfrac{10}{3} $$
Hence, option 'A' is correct.
The value of $$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } $$ is:
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$$0$$
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$$1$$
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$$\displaystyle \frac { \sin { \beta } }{ \beta } $$
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$$\displaystyle \frac { \sin { 2\beta } }{ 2\beta } $$
Explanation
$$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] } $$
$$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] } $$
$$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right) $$ ....... $$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$$
$$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$$ ..... $$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$$
$$=\dfrac { \sin { 2\beta } }{ 2\beta } $$
Evaluate: $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$
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$$\displaystyle \frac{3}{2}$$
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$$\displaystyle \frac{2}{3}$$
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{3}{4}$$
Explanation
Given,
$$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$
$$\sec x=\dfrac{1}{\cos x}$$
$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$$
$$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
$$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
$$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
as $$\cos 0=1$$ and $$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$
$$=\dfrac{3x}{2x}=\dfrac{3}{2}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$$
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$$-\dfrac{1}{2}$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$\dfrac{3}{2}$$
Explanation
We simplify the given expression as,
$$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)} $$
Let $$x - sin x = y$$
As $$ x \rightarrow 0 \text{ so does y } \rightarrow 0 $$
Hence, the question transforms into
$$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y}) $$
$$= \dfrac{1}{2} \times 1 $$
$$= \dfrac{1}{2} $$
$$\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}$$=
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$$1$$
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$$\displaystyle \frac{2}{\pi}$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{5}$$
Explanation
$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x } $$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } } $$
$$\displaystyle =\dfrac{2}{\pi}$$
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$=
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$$\sqrt{2}$$
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$$\displaystyle \frac{1}{\sqrt{2}}$$
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$$-\sqrt{2}$$
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$$\displaystyle \frac{-1}{\sqrt{2}}$$
Explanation
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }$$
$$ =\sqrt { 2 } $$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$=
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$$2$$
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$$-2$$
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$$\displaystyle \frac{1}{2}$$
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$$-\displaystyle \frac{1}{2}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x } }{ 1-\tan ^{ 2 }{ x } } -2x\tan x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x } }{ 1+\tan ^{ 2 }{ x } } \right)^{ 2 } } } $$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}$$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } } $$
$$=\displaystyle \dfrac { 1 }{ 2 } $$
$$\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$$
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$$\sqrt{3}$$
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$$\dfrac{1}{\sqrt{3}}$$
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$$-\sqrt{3}$$
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$$-\dfrac{1}{\sqrt{3}}$$
Explanation
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }$$
$$=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }$$
$$=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=$$
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0
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1
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$$\displaystyle \frac{\pi}{180}$$
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$$\pi$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$$
$$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x } $$
$$\displaystyle={ \dfrac { \pi }{ 180 } }$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$=
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$$0$$
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$$1$$
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$$(\displaystyle \frac{\pi}{180})^{3}$$
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$$4.(\displaystyle \frac{\pi}{180})^{3}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } } $$
$$\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } } $$
$$\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }$$
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