हिंदी

Lim X → a Cos √ X − Cos √ a X − a - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[ = \lim_{x \to a} \frac{- 2 \sin \left( \frac{\sqrt{x} + \sqrt{a}}{2} \right) \sin \left( \frac{\sqrt{x} - \sqrt{a}}{2} \right)}{2\left( \frac{\sqrt{x} - \sqrt{a}}{2} \right) \left( \sqrt{x} + \sqrt{a} \right)}\]
\[ = \lim_{x \to a} \frac{- \sin \left( \frac{\sqrt{x} + \sqrt{a}}{2} \right)}{\sqrt{x} + \sqrt{a}} \times \frac{\sin \left( \frac{\sqrt{x} - \sqrt{a}}{2} \right)}{\left( \frac{\sqrt{x} - \sqrt{a}}{2} \right)}\]
\[ = \frac{- 1 \sin \left( \sqrt{a} \right)}{2\sqrt{a}} \times 1\]
\[ = \frac{- \sin \sqrt{a}}{2\sqrt{a}}\]

 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 16 | पृष्ठ ६२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]


\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 


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


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


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


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


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


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


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


If  \[\lim_{x \to 0} kx  cosec x = \lim_{x \to 0} x  cosec kx,\] 


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


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


\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\] 


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


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


\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]


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


If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\]  then \[\lim_{x \to 0} f\left( x \right) =\] 


\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to


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


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


\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to 


The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is 


Evaluate the following limit:

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


Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`


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


Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when"  x ≠ pi/2),(3",", x = pi/2  "and if"  f(x) = f(pi/2)):}` find the value of k.


Evaluate the following limit.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×