हिंदी

Lim X → 0 Cos 3 X − Cos 7 X X 2

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\] 

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\cos 3x - \cos 7x}{x^2} \right]\] 

\[= \lim_{x \to 0} \left[ \frac{- 2\sin\left( \frac{3x + 7x}{2} \right)\sin\frac{\left( 3x - 7x \right)}{2}}{x^2} \right] \left[ \cos C - \cos D = - 2\sin\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]

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

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

\[ = 2 \lim_{x \to 0} \left[ \frac{\sin 5x}{5x} \times \frac{\sin 2x}{2x} \right] \times 5 \times 2\]

\[\]

\[ = 2 \times 5 \times 2\]

\[ = 20\]

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 15 | पृष्ठ ५०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find `lim_(x -> 5) f(x)`, where f(x)  = |x| - 5


Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


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


\[\lim_{x \to 0} 9\] 


\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\] 


\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\] 


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


\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 


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


\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\] 


\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\] 


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\] 


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


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


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


\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\] 


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


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


\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] 


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


\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\] 


\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\] 


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 


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


\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]


\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]


Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]


\[\lim_{x \to  } \frac{1 - \cos 2x}{x} is\]


\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to


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


Evaluate the following limit:

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


Evaluate the following limit.

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


Evaluate the Following limit: 

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


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×