English

Evaluate the following limit : limx→π[1-cosx-2sin2x]

Advertisements
Advertisements

Question

Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`

Sum
Advertisements

Solution

`lim_(x -> pi) (sqrt(1 - cosx) - sqrt(2))/(sin^2 x)`

= `lim_(x -> pi) (sqrt(1 - cosx) - sqrt(2))/(sin^2x) xx  (sqrt(1 - cosx) + sqrt(2))/(sqrt(1 - cosx) + sqrt(2))`

= `lim_(x -> pi) ((1 - cos x) - 2)/(( 1 - cos^2x)(sqrt(1 - cos x) + sqrt(2))`

= `lim_(x -> pi) (-(1 + cos x))/((1 + cos x)(1 - cosx)(sqrt(1 - cosx) + sqrt(2))`

= `lim_(x -> pi) (-1)/((1 - cosx)(sqrt(1 - cosx) + sqrt(2)))  ...[(because x -> pi","  x ≠ pi),(therefore cos x ≠ cos pi = -1),(therefore 1 + cos x ≠ 0)]`

= `(lim_(x -> pi) (-1))/([lim_(x -> pi) (1 - cosx)] xx [lim_(x -> pi) (sqrt(1 - cosx) + sqrt(2))]`

= `(-1)/((1 - cos pi) (sqrt(1 - cos  pi) + sqrt(2))`

=  `(-1)/((1 + 1)(sqrt(1 + 1) + sqrt(2))`

= `(-1)/(4sqrt(2))`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Limits - Exercise 7.4 [Page 148]

RELATED QUESTIONS

Evaluate the following limit.

`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`


Evaluate the following limit.

`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`


Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`


Evaluate the following limit :

`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`


Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`


Select the correct answer from the given alternatives.

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =


Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


Evaluate the following :

`lim_(x -> 0)[(secx^2 - 1)/x^4]`


Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`


`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______ 


Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`


Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`


Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`


Find the derivative of f(x) = `sqrt(sinx)`, by first principle.


`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.


If f(x) = x sinx, then f" `pi/2` is equal to ______.


Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`


Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`


Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`


Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`


Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`


Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`


Evaluate: `lim_(x -> pi/4)  (sin x - cosx)/(x - pi/4)`


Evaluate: `lim_(x -> 0) (sin 2x + 3x)/(2x + tan 3x)`


Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec"  x - 2)`


cos (x2 + 1)


`(ax + b)/(cx + d)`


`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.


`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.


`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.


`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.


`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.


`lim_(x -> 0) (sin mx cot  x/sqrt(3))` = 2, then m = ______. 


If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.


`lim_(theta → -pi/4) (cos theta + sin theta)/(theta + pi/4)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×