हिंदी

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

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

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

योग
Advertisements

उत्तर

`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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.4 [पृष्ठ १४८]

APPEARS IN

संबंधित प्रश्न

Evaluate the following limit.

`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`


Evaluate the following limit.

`lim_(x -> 0) (ax +  xcos x)/(b sin x)`


Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`


Evaluate the following limit :

`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`


Evaluate the following limit :

`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`


Select the correct answer from the given alternatives.

`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =


Select the correct answer from the given alternatives.

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


Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`


Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.


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


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


Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`


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


Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x  - 1)`


Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`


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


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


Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`


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


Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists


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


`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.


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


`lim_(x -> 3^+) x/([x])` = ______.


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.


`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.


`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×