मराठी

Limx→0tan2x-x3x-sinx is equal to ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • 2

  • `1/2`

  • `-1/2`

  • `1/4`

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to `1/2`.

Explanation:

Given `lim_(x -> 0) (tan 2x - x)/(3x - sin x)`

= `lim_(x -> 0) (x[tan2x/x - 1])/(x[3 - sin x/x])`

`lim_((x -> 0),(because  2x -> 0)) ((tan 2x)/(2x) xx 2 - 1)/(3 - sinx/x)`

= `(1.2 - 1)/(3 - 1)`

= `(2 - 1)/2`

= `1/2`

∴ 2x → 0

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 66 | पृष्ठ २४४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


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 the following :

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


`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______ 


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


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


`lim_(x -> 0) |x|/x` is equal to ______.


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


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


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


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


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


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


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


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


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


`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`


`lim_(x -> pi) sinx/(x - pi)` is equal to ______.


`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.


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


`lim_(x -> 0) |sinx|/x` is ______.


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


If L = `lim_(x→∞)(x^2sin  1/x - x)/(1 - |x|)`, then value of L is ______.


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×