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

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

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

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

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

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

$f(x) = sin x$ , find

$\frac{dy}{dx}$ .

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

Explanation

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

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

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

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

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

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

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

Differentiate:

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

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

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

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

Explanation

We have,

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

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

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

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

$=\dfrac{7}{3}$

Hence, this is the answer.

Consider the differential equation

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

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

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

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

$g(x)\ne 0$ 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 :

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

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

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

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

Explanation

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

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

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

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

$\dfrac {dy}{dx}$

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

Explanation

The Equation is

$y=5x^2+8x$

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

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

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

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%
$none\ of\ these$

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

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

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%
$none\ of\ these$

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

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

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

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

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

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

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

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

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

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

$a,b$ respectively are

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

Explanation

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

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

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

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

$\Rightarrow 1+a+b=0$

$\Rightarrow a+b=-1$

Now, put

$a = -4$

$\Rightarrow b = 3$

Option C satisfies above equation.

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

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

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

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

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

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

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

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

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

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

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

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 :

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

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

Explanation

Solve :

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

$=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)$

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

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

$=\dfrac34$

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

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

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:

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

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:

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

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

Explanation

Given,

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

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

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$

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

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

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

$\cos 0=1$ and

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

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

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

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

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

$x - sin x = y$

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

Hence, the question transforms into

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

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

$= \dfrac{1}{2}$

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

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

Explanation

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

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

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

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

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

Explanation

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

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

$=\sqrt { 2 }$

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

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

Explanation

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

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

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

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

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

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

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

Explanation

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

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

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

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

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

Explanation

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

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

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

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

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

Explanation

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

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

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

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

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

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

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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