Find $$\dfrac{{dy}}{{dx}}$$ of the following $$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$$

$$x+2x^2+6x^3+(n-1)(n-2)x^{n-3}$$

Consider the differential equation $$\frac { d y } { d x } = \cos x$$ Then we observe that

$$y = \sin x - \frac { 1 } { 2 }$$

Evaluate $$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$

$$\cfrac { 1 }{ { 2 }^{ 2 } } $$

If $$y=e^{\sin \sqrt x}$$ then $$\dfrac{dy}{dx}=?$$

$$\dfrac{e^{\sin \sqrt x}\cos \sqrt x}{2\sqrt x}$$

$$e^{\sin \sqrt x}.\cos \sqrt x$$

$$\dfrac{e^{\sin \sqrt x}}{2\sqrt x}$$

If $$y=x^2\sin \dfrac{1}{x}$$ then $$\dfrac{dy}{dx}=?$$

$$-\cos \dfrac{1}{x}+2x\sin \dfrac{1}{x}$$

$$x\sin \dfrac{1}{x}-\cos \dfrac{1}{x}$$

$$-x\sin \dfrac{1}{x}+\cos \dfrac{1}{x}$$

If $$(x+y)=\sin (x+y)$$ then $$\dfrac{dy}{dx}=?$$

$$\dfrac{1-\cos (x+y)}{\cos^2(x+y)}$$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 2

Find

\frac{d y}{d x}

$\dfrac{{dy}}{{dx}}$ of the following

y = 1 + 2 x + 3 x^{2} + (n - 1) x^{n - 2}

$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$

0%
$x+2x^2+6x^3+(n-1)(n-2)x^{n-3}$
0%
$2+6x+(n-1)x^{n-2}$
0%
$2+6x+(n-1)(n-2)x^{n-3}$
0%
None

Explanation

y = 1 + 2 x + 3 x^{2} + (n - 1) x^{n - 2}

$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$

\frac{d y}{d x} = 2 + 6 x + (n - 1) (n - 2) x^{n - 3}

$\dfrac{{dy}}{{dx}} = 2 + 6x + \left( {n - 1} \right)\left( {n - 2} \right){x^{n - 3}}$

f (x) = s i n x

$f(x) = sin x$ , find

\frac{d y}{d x}

$\frac{dy}{dx}$ .

0%
$cosx$
0%
$-cosx$
0%
$cotx$
0%
$-cot^2x$

Explanation

y = sin x \Rightarrow \frac{d y}{d x} = cos x

$y=\sin x\Rightarrow \dfrac{dy}{dx}=\cos x$

x^{\frac{1}{2}} + 1 = t

$x^{\frac{1}{2}} + 1= t$
differentiate w.r.t. x

0%
$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{4\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{8\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{16\sqrt{x}}$

Explanation

x^{\frac{1}{2}} + 1 = t

$x^{\frac{1}{2}} + 1= t$

\frac{d t}{d x} = \frac{1}{2} (x)^{\frac{1}{2} - 1} + 0

$\dfrac{dt}{dx} = \dfrac{1}{2}(x) ^{\frac{1}{2}-1} + 0$

\frac{1}{2} (x)^{\frac{- 1}{2}}

$\frac{1}{2}(x) ^{\frac{-1}{2}}$

\frac{1}{2 x^{\frac{1}{2}}}

$\dfrac{1}{2x^{\frac{1}{2}}}$

\frac{d t}{d x} = \frac{1}{2 \sqrt{x}}

$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$

Differentiate:

x^{100} + sin x - 1

$x^{100} + \sin x - 1$

0%
$100x^{99}-\cos x$
0%
$100x^{99}+\cos x$
0%
$x^{99}+\cos x$
0%
$100x^{99}+\sin x$

Explanation

\frac{d}{d x} (x^{100} + sin x - 1)

$\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} +\sin x -1 \right )$

= \frac{d}{d x} (x^{100}) + \frac{d}{d x} (sin x) - \frac{d}{d x} (1)

$=\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} \right )+\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( \sin x \right )-\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( 1 \right )$

= 100 x^{100 - 1} + cos x - 0

$=100x^{100-1}+\cos x -0$

= 100 x^{99} + cos x

$=100x^{99}+\cos x$

lim x \to 0 \frac{sin 7 x}{sin 3 x}

$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$ equals

0%
$\dfrac{7}{3}$
0%
$\dfrac{10}{3}$
0%
$\dfrac{14}{3}$
0%
$\dfrac{1}{3}$

Explanation

We have,

lim x \to 0 \frac{sin 7 x}{sin 3 x}

$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$

lim x \to 0 \frac{sin 7 x}{sin 3 x}

$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$

lim x \to 0 \frac{\frac{sin 7 x}{7 x} \times 7 x}{\frac{sin 3 x}{3 x} \times 3 x}

$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7x}{\dfrac{\sin3x}{3x}\times 3x}$

lim x \to 0 \frac{\frac{sin 7 x}{7 x} \times 7}{\frac{sin 3 x}{3 x} \times 3}

$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7}{\dfrac{\sin3x}{3x}\times 3}$

= \frac{1 \times 7}{1 \times 3}

$=\dfrac{1\times 7}{1\times 3}$

= \frac{7}{3}

$=\dfrac{7}{3}$

Hence, this is the answer.

Consider the differential equation

\frac{d y}{d x} = cos x

$\frac { d y } { d x } = \cos x$ Then we observe that

0%
$y = \sin x$
0%
$y = \sin x + 2$
0%
$y = \sin x - \frac { 1 } { 2 }$
0%
$y = \sin x + c$

Explanation

d y = cos x d x \int d y = \int cos x d x y = sin x + C

$dy = \cos x\>dx\\\int dy=\int \cos x\>dx\\y=\sin x+C$

We must add constant "C" as its family of curves.

Which of the following statement is not correct

0%
$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) }$
0%
$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } }$

Explanation

All the result other than D are correct.

The last result will hold only when

g (x) \neq 0

$g(x)\ne 0$ and

lim x \to c g (x) \neq 0

$\lim\limits_{x\to c} g(x)\ne 0$ for

x

$x$ belonging to a neighbourhood of

c

$c$ .

A, B, C are general results which are always true.

Evaluate the following limit :

l i m_{x \to 0} \frac{1 - cos 2 x}{x^{2}}

$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$

0%
$0$
0%
$1$
0%
$2$
0%
none of these

Explanation

l i m_{x \to 0} \frac{1 - cos 2 x}{x^{2}}

$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$

= l i m_{x \to 0} \frac{2 {sin}^{2} x}{x^{2}}

$=lim_{x\rightarrow 0} \dfrac{2\sin^2 x}{x^2}$

= 2 l i m_{x \to 0} (\frac{sin x}{x} \times \frac{sin x}{x})

$=2lim_{x \rightarrow 0} (\dfrac{\sin x}{x} \times \dfrac{\sin x}{x})$

= 2 l i m_{x \to 0} \frac{sin x}{x} \times l i m_{x \to 0} \frac{sin x}{x} = 2 (1) (1) = 2

$=2lim_{x\rightarrow 0} \dfrac{\sin x}{x} \times lim_{x\rightarrow 0} \dfrac{\sin x}{x}=2(1)(1)=2$

Evaluate

lim n \to \infty \frac{1 + 2 + 3 + . . . + n}{n^{2}}

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$

0%
$\cfrac { 1 }{ 2 }$
0%
$1$
0%
${ 3 }^{ 2 }$
0%
$\cfrac { 1 }{ { 2 }^{ 2 } }$

Explanation

lim n \to \infty \frac{1 + 2 + 3 + . . . + n}{n^{2}}

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$

= lim n \to \infty \frac{1}{n^{2}} \times \frac{n (n + 1)}{2}

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

= lim n \to \infty \frac{1}{2} (1 + \frac{1}{n}) = \frac{1}{2}

$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{2}(1+\dfrac{1}{n})=\dfrac{1}{2}$

y = 5 x^{2} + 8 x

$y=5x^2+8x$ find

\frac{d y}{d x}

$\dfrac {dy}{dx}$

0%
$10x+8$
0%
$5x+8$
0%
$10x^2+8x$
0%
none of these

Explanation

The Equation is

y = 5 x^{2} + 8 x

$y=5x^2+8x$

\frac{d y}{d x} = \frac{d}{d x} (5 x^{2} + 8 x) \frac{d y}{d x} = 10 x + 8

$\dfrac {dy}{dx}=\dfrac d{dx}(5x^2+8x)\\\dfrac{dy}{dx}=10x+8$

y = e^{sin \sqrt{x}}

$y=e^{\sin \sqrt x}$ then

\frac{d y}{d x} = ?

$\dfrac{dy}{dx}=?$

0%
$e^{\sin \sqrt x}.\cos \sqrt x$
0%
$\dfrac{e^{\sin \sqrt x}\cos \sqrt x}{2\sqrt x}$
0%
$\dfrac{e^{\sin \sqrt x}}{2\sqrt x}$
0%
$none\ of\ these$

y = x^{2} sin \frac{1}{x}

$y=x^2\sin \dfrac{1}{x}$ then

\frac{d y}{d x} = ?

$\dfrac{dy}{dx}=?$

0%
$x\sin \dfrac{1}{x}-\cos \dfrac{1}{x}$
0%
$-\cos \dfrac{1}{x}+2x\sin \dfrac{1}{x}$
0%
$-x\sin \dfrac{1}{x}+\cos \dfrac{1}{x}$
0%
$none\ of\ these$

(x + y) = sin (x + y)

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

\frac{d y}{d x} = ?

$\dfrac{dy}{dx}=?$

0%
$-1$
0%
$1$
0%
$\dfrac{1-\cos (x+y)}{\cos^2(x+y)}$
0%
$none\ of\ these$

x = a {cos}^{2} θ, y = b {sin}^{2} θ

$x=a\cos^{2}\theta, y=b\sin^{2}\theta$ then

\frac{d y}{d x} = ?

$\dfrac{dy}{dx}=?$

0%
$\dfrac{-a}{b}$
0%
$\dfrac{a}{b}\cot \theta$
0%
$\dfrac{-b}{a}$
0%
$none\ of\ these$

y = {cos}^{2} x^{3}

$y=\cos^2 x^3$ then

\frac{d y}{d x} = ?

$\dfrac{dy}{dx}=?$

0%
$-3x^2\sin (2x^3)$
0%
$-3x^2\sin^2x^3$
0%
$-3x^2\cos^2(2x^3)$
0%
$none\ of\ these$

lim x \to 0 (\frac{cos 4 x + a cos 2 x + b}{x^{4}})

$\displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)$ is finite then the value of

a, b

$a,b$ respectively are

0%
$5\, -4$
0%
$-5,\ -4$
0%
$-4,\ 3$
0%
$4,\ 5$

Explanation

lim x \to 0 (\frac{cos 4 x + a cos 2 x + b}{x^{4}})

$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)$

x \to 0

${ x\rightarrow 0 }$ , denominator tends to 0, so the numerator also tends to 0.

\Rightarrow lim x \to 0 cos 4 x + a cos 2 x + b = 0

$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$

x = 0 \Rightarrow cos 4 x = 1, cos 2 x = 1

$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$

\Rightarrow 1 + a + b = 0

$\Rightarrow 1+a+b=0$

\Rightarrow a + b = - 1

$\Rightarrow a+b=-1$

Now, put

a = - 4

$a = -4$

\Rightarrow b = 3

$\Rightarrow b = 3$

Option C satisfies above equation.

lim x \to 1 (1 - x) tan (\frac{π x}{2}) =

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$

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

Explanation

lim x \to 1 (1 - x) tan (\frac{π x}{2})

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$

= lim x \to 1 \frac{1 - x}{cot (\frac{π x}{2})}

$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) }$
It is of the form

\frac{0}{0}

$\displaystyle \dfrac{0}{0}$ , so applying L-Hospital's rule

= lim x \to 1 \frac{- 1}{- \frac{π}{2} {csc}^{2} (\frac{π x}{2})}

$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } }$

= \frac{2}{π}

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

lim x \to 0 \frac{1 - {cos}^{3} x}{x sin 2 x}

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

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

Explanation

lim x \to 0 \frac{1 - {cos}^{3} x}{x sin 2 x} = lim x \to 0 \frac{(1 - cos x) (1 + cos x + {cos}^{2} x)}{x sin 2 x}

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

= lim x \to 0 \frac{2 {sin}^{2} \frac{x}{2} (1 + cos x + {cos}^{2} x)}{4 x cos x sin \frac{x}{2} cos \frac{x}{2}} = lim x \to 0 \frac{sin \frac{x}{2} (1 + cos x + {cos}^{2} x)}{4 \times \frac{x}{2} cos \frac{x}{2} \times cos x} = \frac{3}{4}

$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$ ...... As

⎧ ⎨ ⎩ lim \frac{x}{2} \to 0 \frac{sin \frac{x}{2}}{\frac{x}{2}} = 0 ⎫ ⎬ ⎭

$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$

lim x \to 0 \frac{1 - cos x}{x log (1 + x)} =

$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$

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

Explanation

lim x \to 0 \frac{1 - cos x}{x log (1 + x)}

$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$

lim x \to 0 \frac{2 {sin}^{2} \frac{x}{2}}{x log (1 + x)}

$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$

= lim x \to 0 \frac{2 \frac{{sin}^{2} \frac{x}{2}}{{(\frac{x}{2})}^{2}} \times \frac{1}{4}}{\frac{1}{x} log (1 + x)}

$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) }$

= lim x \to 0 \frac{\frac{1}{2}}{\frac{log (1 + x)}{x}} = \frac{1}{2}

$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$

Solve :

lim x \to 0 \frac{sin x sin (\frac{π}{3} + x) sin (\frac{π}{3} - x)}{x}

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$

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

Explanation

Solve :

lim x \to 0 \frac{sin x sin (\frac{π}{3} + x) sin (\frac{π}{3} - x)}{x}

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$

= lim x \to 0 \frac{sin x}{x} \times lim x \to 0 sin (\frac{π}{3} + x) \times lim x \to 0 sin (\frac{π}{3} - 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 \frac{π}{3} \times sin \frac{π}{3}

$=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}$

= 1 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}

$=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}$

= \frac{3}{4}

$=\dfrac34$

lim x \to 0 \frac{(1 - cos 2 x) sin 5 x}{x^{2} sin 3 x} =

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

0%
$\dfrac{10}{3}$
0%
$\dfrac{3}{10}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{5}{6}$

Explanation

Using,

1 - c o s 2 x = 2 {s i n}^{2} x

$1 - cos 2x = 2{sin}^{2} x$ ,
The expression transforms to,

l i m_{x \to 0} \frac{2 {s i n}^{2} x s i n 5 x}{x^{2} s i n 3 x}

$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x}$
Rewriting the expression in a different form,

l i m_{x \to 0} \frac{2 {s i n}^{2} x}{x^{2}} \times \frac{s i n 5 x}{5 x} \times \frac{3 x}{s i n 3 x} \times \frac{5}{3}

$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3}$
Therefore, the limit to the expression is

\frac{10}{3}

$\dfrac{10}{3}$
Hence, option 'A' is correct.

The value of

lim α \to β [\frac{{sin}^{2} α - {sin}^{2} β}{α^{2} - β^{2}}]

$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] }$ is:

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

Explanation

lim α \to β [\frac{{sin}^{2} α - {sin}^{2} β}{α^{2} - β^{2}}] = lim α \to β [\frac{sin (α - β) sin (α + β)}{(α - β) (α + β)}]

$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] }$

= lim α \to β [\frac{sin (α - β)}{(α - β)}] \times lim α \to β [\frac{sin (α + β)}{(α + β)}]

$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] }$

= lim α - β \to 0 [\frac{sin (α - β)}{(α - β)}] . (\frac{sin 2 β}{2 β})

$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right)$ .......

[∵ α \to β ⟹ α - β \to 0]

$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$

= 1. \frac{sin 2 β}{2 β}

$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$ .....

[∵ lim x \to 0 \frac{sin x}{x} = 1]

$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$

= \frac{sin 2 β}{2 β}

$=\dfrac { \sin { 2\beta } }{ 2\beta }$

Evaluate:

lim x \to 0 \frac{sec 4 x - sec 2 x}{sec 3 x - sec x}

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

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

Explanation

Given,

lim x \to 0 \frac{sec 4 x - sec 2 x}{sec 3 x - sec x}

$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

sec x = \frac{1}{cos x}

$\sec x=\dfrac{1}{\cos x}$

= lim x \to 0 \frac{\frac{1}{cos 4 x} - \frac{1}{cos 2 x}}{\frac{1}{cos 3 x} - \frac{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 \to 0 \frac{cos 2 x - cos 4 x}{cos x - cos 3 x} \frac{cos 3 x cos x}{cos 4 x cos 2 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 \to 0 \frac{- 2 sin 3 x sin (- x)}{- 2 sin (2 x) sin (- x)} \frac{cos 3 x cos x}{cos 4 x cos 2 x}

$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$

= lim x \to 0 \frac{\frac{sin 3 x}{3 x} \times 3 x}{\frac{sin 2 x}{2 x} \times 2 x} \frac{cos 3 x cos x}{cos 4 x cos 2 x}

$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$

cos 0 = 1

$\cos 0=1$ and

lim x \to 0 \frac{sin x}{x} = 1

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

= \frac{3 x}{2 x} = \frac{3}{2}

$=\dfrac{3x}{2x}=\dfrac{3}{2}$

lim x \to 0 \frac{e^{x} - e^{sin x}}{2 (x - sin x)} =

$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$

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

Explanation

We simplify the given expression as,

lim x \to 0 \frac{e^{s i n x} (e^{x - s i n x} - 1)}{2 (x - s i n x)}

$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)}$

Let

x - s i n x = y

$x - sin x = y$

x \to 0 so does y \to 0

$x \rightarrow 0 \text{ so does y } \rightarrow 0$

Hence, the question transforms into

(lim x \to 0 \frac{e^{s i n x}}{2}) \times (lim y \to 0 \frac{e^{y} - 1}{y})

$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y})$

= \frac{1}{2} \times 1

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

= \frac{1}{2}

$= \dfrac{1}{2}$

lim x \to \frac{π}{2} \frac{c o s e c x - cot x}{x}

$\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}$ =

0%
$1$
0%
$\displaystyle \frac{2}{\pi}$
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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{5}$

Explanation

lim x \to \frac{π}{2} \frac{csc x - cot x}{x}

$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x }$

= lim x \to \frac{π}{2} \frac{1 - cos x}{x sin x}

$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } }$

= \frac{2}{π}

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

lim x \to \frac{π}{4} \frac{sec x . tan (4 x - π)}{sin (4 x - π)}

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$ =

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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

lim x \to \frac{π}{4} \frac{sec x . tan (4 x - π)}{sin (4 x - π)}

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$

= lim x \to \frac{π}{4} \frac{sec x . \frac{tan (4 x - π)}{(4 x - π)}}{\frac{sin (4 x - π)}{(4 x - π)}}

$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }$

= \sqrt{2}

$=\sqrt { 2 }$

lim x \to 0 \frac{x tan 2 x - 2 x tan x}{(1 - cos 2 x)^{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

lim x \to 0 \frac{x tan 2 x - 2 x tan x}{(1 - cos 2 x)^{2}}

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

= lim x \to 0 \frac{x \frac{2 tan x}{1 - {tan}^{2} x} - 2 x tan x}{{(1 - \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x})}^{2}}

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

= lim x \to 0 \frac{2 x {tan}^{3} x}{4 {tan}^{4} x}

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}$

= lim x \to 0 \frac{2 x}{4 tan x}

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } }$

= \frac{1}{2}

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

lim x \to \frac{π}{6} \frac{3 sin x - \sqrt{3} cos x}{6 x - π} =

$\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$

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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

lim x \to \frac{π}{6} \frac{3 sin x - \sqrt{3} cos x}{6 x - π} = 2 \sqrt{3} lim x \to \frac{π}{6} \frac{\frac{\sqrt{3}}{2} sin x - \frac{1}{2} cos x}{6 x - π}

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }$

= \frac{2 \sqrt{3}}{6} lim x \to \frac{π}{6} \frac{cos \frac{π}{6} sin x - sin \frac{π}{6} cos x}{x - \frac{π}{6}}

$=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }$

= \frac{1}{\sqrt{3}} lim x \to \frac{π}{6} \frac{sin (x - \frac{π}{6})}{(x - \frac{π}{6})} = \frac{1}{\sqrt{3}}

$=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}$

lim x \to 0 \frac{t a n x^{0}}{x} =

$\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=$

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0
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1
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$\displaystyle \frac{\pi}{180}$
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$\pi$

Explanation

lim x \to 0 \frac{t a n x^{0}}{x}

$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$

= \frac{π}{180} lim x \to 0 \frac{tan (\frac{π}{180} x)}{(\frac{π}{180}) x}

$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x }$

= \frac{π}{180}

$\displaystyle={ \dfrac { \pi }{ 180 } }$

lim x \to 0 \frac{3 sin x - sin 3 x}{x^{3}}

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$ =

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$0$
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$1$
0%
$(\displaystyle \frac{\pi}{180})^{3}$
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$4.(\displaystyle \frac{\pi}{180})^{3}$

Explanation

lim x \to 0 \frac{3 sin x - sin 3 x}{x^{3}}

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$

= lim x \to 0 \frac{4 {sin}^{3} x}{x^{3}}

$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }$

= lim x \to 0 \frac{4 {sin}^{3} (\frac{π}{180} x)}{x^{3}}

$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } }$

= {(\frac{π}{180})}^{3} lim x \to 0 \frac{4 {sin}^{3} (\frac{π}{180} x)}{x^{3} {(\frac{π}{180})}^{3}}

$\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } }$

= 4 {(\frac{π}{180})}^{3}

$\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 2 - 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 2 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

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