हिंदी

Evaluate the following limit. limx→πsin(π-x)π(π-x)

Advertisements
Advertisements

प्रश्न

Evaluate the following limit.

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

योग
Advertisements

उत्तर

`lim_(x → π)(sin(π - x))/(π (π - x))`

Let π − x = θ, When x → π, θ → 0

∴ `lim_(x → π) 1/π. (sin(π - x))/((π - x))`

= `lim_(θ → 0) (sinθ)/(πθ)`

= `lim_(θ → 0) 1/π. ((sin θ)/θ)`

= `1/π`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Limits and Derivatives - EXERCISE 12.1 [पृष्ठ २३८]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
अध्याय 12 Limits and Derivatives
EXERCISE 12.1 | Q 15. | पृष्ठ २३८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate the following limit.

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 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 -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =


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


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


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


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


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


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


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


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


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


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


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 -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`


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


cos (x2 + 1)


`x^(2/3)`


`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`


`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.


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


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


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


If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______. 


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


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 ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×