मराठी

Evaluate the Following Limit: Lim X → π 1 − Sin X 2 Cos X 2 ( Cos X 4 − Sin X 4 ) - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 

Advertisements

उत्तर

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

Put

\[x = \pi + h\] When \[x \to \pi, h \to 0\] 

\[\therefore \lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \sin\left( \frac{\pi + h}{2} \right)}{\cos\left( \frac{\pi + h}{2} \right)\left[ \cos\left( \frac{\pi + h}{4} \right) - \sin\left( \frac{\pi + h}{4} \right) \right]}\]
\[ = \lim_{h \to 0} \frac{1 - \sin\left( \frac{\pi}{2} + \frac{h}{2} \right)}{\cos\left( \frac{\pi}{2} + \frac{h}{2} \right)\left[ \cos\left( \frac{\pi}{4} + \frac{h}{4} \right) - \sin\left( \frac{\pi}{4} + \frac{h}{4} \right) \right]}\]
\[ = \lim_{h \to 0} \frac{1 - \cos\left( \frac{h}{2} \right)}{- \sin\left( \frac{h}{2} \right)\left[ \left( \cos\frac{\pi}{4}\cos\frac{h}{4} - \sin\frac{\pi}{4}\sin\frac{h}{4} \right) - \left( \sin\frac{\pi}{4}\cos\frac{h}{4} + \cos\frac{\pi}{4}\sin\frac{h}{4} \right) \right]}\]

\[= \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{4}}{- 2\sin\frac{h}{4}\cos\frac{h}{4}\left( \frac{1}{\sqrt{2}}\cos\frac{h}{4} - \frac{1}{\sqrt{2}}\sin\frac{h}{4} - \frac{1}{\sqrt{2}}\cos\frac{h}{4} - \frac{1}{\sqrt{2}}\sin\frac{h}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{\sin\frac{h}{4}}{- \cos\frac{h}{4} \times \left( - \sqrt{2}\sin\frac{h}{4} \right)}\]
\[ = \frac{1}{\sqrt{2}} \times \frac{1}{\lim_{h \to 0} \cos\frac{h}{4}}\]
\[ = \frac{1}{\sqrt{2}} \times 1 \left( \lim_\theta \to 0 \cos\theta = 1 \right)\]
\[ = \frac{1}{\sqrt{2}}\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.8 [पृष्ठ ६३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.8 | Q 38 | पृष्ठ ६३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


\[\lim_{x \to 0} 9\] 


\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]


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


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


\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\] 


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


\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\] 


\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\] 


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


\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]


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


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


\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\] 


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


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


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


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


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


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


\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]


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


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


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


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


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


If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]


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


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


Evaluate the Following limit:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×