English

Lim X → π 2 ( π 2 − X ) Tan X

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[ = \lim_{x \to 1} \frac{x - 1}{x \sin \pi\left( x - 1 \right)}\]
\[Let y = x - 1\]
\[ x \to 1\]
\[ \therefore y \to 0\]
\[ = \lim_{y \to 0} \frac{y}{\left( y + 1 \right) \sin \pi y}\]
\[ = \lim_{y \to 0} \frac{1}{\pi\left( y + 1 \right) \times \frac{\sin \pi y}{\pi y}}\]
\[ = \frac{1}{\pi\left( 0 + 1 \right) \times 1}\]
\[ = \frac{1}{\pi}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.8 [Page 63]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.8 | Q 33 | Page 63

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


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


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


\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]


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


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


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


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


Evaluate the following limit:

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


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


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


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


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


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


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


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


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


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


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


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


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


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


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


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


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


Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\] 


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to 


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 


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


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


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_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.


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.


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


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×