हिंदी

Lim X → 0 Sin 5 X − Sin 3 X Sin X

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\] 

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin 5x - \sin 3x}{\sin x} \right]\]

\[= \lim_{x \to 0} \left[ \frac{2\cos\left( \frac{5x + 3x}{2} \right)\sin\left( \frac{5x - 3x}{2} \right)}{\sin x} \right] \left[ \because sinC - sinD = 2\cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2\cos4x . \sin x}{\sin x} \right]\]
\[ = 2 \cos0\]
\[ = 2\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५०]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 23 | पृष्ठ ५०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\] 


\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\] 


\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\] 


If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a


\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\] 


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


\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\] 


\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 


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


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


\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\] 


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


\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\] 


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\] 


\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]


\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\] 


\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]


\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]


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


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


Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]

 

Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]


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


Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]


\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]


\[\lim_{x \to 0} \frac{x}{\tan x} is\] 


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\] 


\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to 


\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to 


\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]


Evaluate the following Limit:

`lim_(x -> 0) ((1 + x)^"n" - 1)/x`


Which of the following function is not continuous at x = 0?


If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on ______.


Evaluate the following limit:

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


Evaluate the following limit.

`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×