हिंदी

Evaluate: limx→π63sinx-cosxx-π6

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Given that `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`

= `lim_(x -> pi/6) (2[sqrt(3)/2  sin x - 1/2 cos x])/(x - pi/6)`

= `lim_(x -> pi/6) (2[cos  pi/6  sin x - sin  pi/6  cos x])/(x - pi/6)`

= `lim_((x -> pi/6),(because  x - pi/6 -> 0)) (2sin (x - pi/6))/((x - pi/6))`  .......`[because  lim_(x -> 0) sinx.x = 1]`

= `2 * 1`

= 2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 22 | पृष्ठ २४०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate the following limit.

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


Evaluate the following limit.

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


Evaluate the following limit :

`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`


Evaluate the following limit :

`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`


Evaluate the following limit :

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


Evaluate the following :

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


Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`


Evaluate the following :

`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`


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


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


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


`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.


`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.


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


Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`


Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`


Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`


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


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


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


Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`


`x^(2/3)`


`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`


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


`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.


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


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


If `lim_(n→∞)sum_(k = 2)^ncos^-1(1 + sqrt((k - 1)(k + 2)(k + 1)k)/(k(k + 1))) = π/λ`, then the value of λ is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×