मराठी

Lim X → π 4 √ Cos X − √ Sin X X − π 4 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\text{ Rationalising the numerator }: \]
\[ = \lim_{x \to \frac{\pi}{4}} \left( \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}} \right) \left( \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right)\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( x - \frac{\pi}{4} \right) \left( \sqrt{\cos x} + \sqrt{\sin x} \right)}\]
\[ = \lim_{h \to 0} \frac{\cos \left( \frac{\pi}{4} + h \right) - \sin \left( \frac{\pi}{4} + h \right)}{\left( \frac{\pi}{4} + h - \frac{\pi}{4} \right) \left( \sqrt{\cos \left( \frac{\pi}{4} + h \right)} + \sqrt{\sin \left( \frac{\pi}{4} + h \right)} \right)}\]
\[ = \lim_{h \to 0} \frac{\cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h - \sin \frac{\pi}{4} \cos h - \cos \frac{\pi}{4} \sin h}{h\left( \sqrt{\cos \left( \frac{\pi}{4} + h \right)} + \sqrt{\sin \left( \frac{\pi}{4} + h \right)} \right)}\]
\[ = \lim_{h \to 0} \frac{- \sqrt{2} \sin h}{h\left( \sqrt{\cos \left( \frac{\pi}{4} + h \right)} + \sqrt{\sin \left( \frac{\pi}{4} + h \right)} \right)}\]
\[ \Rightarrow \frac{- \sqrt{2} \times 1}{2 \left( \frac{1}{2} \right)^\frac{1}{4}}\]
\[ = \frac{- 1}{2^\frac{1}{4}} = - 2^{- \frac{1}{4}}\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.8 [पृष्ठ ६२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.8 | Q 10 | पृष्ठ ६२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\] 


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


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


\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 


\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\] 


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


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


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


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


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


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


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


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


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


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


\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\] 


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


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)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×