मराठी

Lim X → 0 3 Sin X − 4 Sin 3 X X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{3\sin x - 4 \sin^3 x}{x} \right]\] 

=  \[\lim_{x \to 0} \left[ \frac{\sin 3x}{x} \right]\] \[\left[ \because \sin^3 A = 3sinA - 4 \sin^3 A \right]\]

=  \[\lim_{x \to 0} \left[ \frac{\sin 3x}{3x} \times 3 \right]\] 

= 1 × 3

= 3

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.7 [पृष्ठ ५०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.7 | Q 5 | पृष्ठ ५०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\] 


\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\] 


\[\lim_{x \to 2} \left( 3 - x \right)\] 


\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\] 


\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\] 


\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]


Evaluate the following limit:

\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\] 


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]


\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\] 


\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 


\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\] 


\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\] 


\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\] 


\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\] 


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\] 


\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]


\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]


\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]


\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\] 


\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]


\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]


Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]


\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\] 


\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\] 


\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is 


The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is 


Evaluate the following limit:

`lim_(x -> 3) [sqrt(x + 6)/x]`


`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?


Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`


Evaluate the following limit:

`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×