हिंदी

The Value of Lim X → π / 2 ( Sec X − Tan X ) is - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • 2

  • −1

  •  1

  • 0

MCQ

उत्तर

\[\lim_{x \to \frac{\pi}{2}} \left( \sec x - \tan x \right)\]
\[ = \lim_{h \to 0} \left( \sec \left( \frac{\pi}{2} - h \right) - \tan \left( \frac{\pi}{2} - h \right) \right)\]
\[ = \lim_{h \to 0} \left( cosec h - \cot h \right)\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{\sin h}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{\sin h}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{2 \sin \frac{h}{2}\cos \frac{h}{2}}\]
\[ = \lim_{h \to 0} \tan \frac{h}{2}\]
\[ = 0\]

 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ८०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 34 | पृष्ठ ८०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]


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


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


\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\] 


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


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


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


\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\] 


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


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


\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]


\[\lim_{x \to 0} \frac{x}{\tan x} is\] 


\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals 


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to 


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


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 -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


`1/(ax^2 + bx + c)`


Evaluate the Following limit:

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


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)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×