हिंदी

Lim X → 3 π 2 1 + C O S E C 3 X Cot 2 X

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]

Advertisements

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.9 [पृष्ठ ६५]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.9 | Q 6 | पृष्ठ ६५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\] 


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


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


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


\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\] 


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]


\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\] 


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


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


Evaluate the following limit: 

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]


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


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


\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\] 


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


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


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


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


Evaluate the following limit:

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

 


\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 


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


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


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 \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is 


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.


Evaluate the following limit :

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


Evaluate the following limit:

`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


Evaluate the following limit.

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


Evaluate the Following limit: 

`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)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×