हिंदी

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

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[ = \lim_{x \to a} \frac{- 2 \sin \left( \frac{x + a}{2} \right) \sin \left( \frac{x - a}{2} \right)}{\sqrt{x} - \sqrt{a}}\]
\[\text{ Dividing the numerator and the denominator by  }\sqrt{x} + \sqrt{a}: \]
\[ - 2 \lim_{x \to a} \frac{\left( \sin \left( \frac{x + a}{2} \right)\sin \left( \frac{x - a}{2} \right) \right)}{2\left( \frac{x - a}{2} \right)}\left( \sqrt{x} + \sqrt{a} \right)\]
\[ = - 2 \lim_{x \to a} \sin \left( \frac{x + a}{2} \right) \times \frac{\sin \left( \frac{x - a}{2} \right)}{2\left( \frac{x - a}{2} \right)}\left( \sqrt{x} + \sqrt{a} \right)\]
\[ \Rightarrow - 2 \sin \left( \frac{2a}{2} \right) \left( \frac{\sqrt{a} + \sqrt{a}}{2} \right)\]
\[ = - 2\sqrt{a} \sin a\]

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 14 | पृष्ठ ६२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\] 


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


\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\] 


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


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


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


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


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


\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]


\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]


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


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


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


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


Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\] 


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


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


\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to 


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


If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\]  where  denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]  


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


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


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 limits: `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)]`


Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×