English

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

Advertisements
Advertisements

Question

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

Solution

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

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
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.8 [Page 63]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.8 | Q 38 | Page 63

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


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


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


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


\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\] 


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


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


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


\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\] 


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\] 


Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\] 


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


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


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


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


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


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


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


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


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


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


\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]


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


If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 


\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to 


The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\] 


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


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


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


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`


Evaluate the following limit:

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


Evaluate the following limit:

`\underset{x->3}{lim}[sqrt(x +6)/(x)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×