मराठी

Limx→0(sinmxcot x3) = 2, then m = ______.

Advertisements
Advertisements

प्रश्न

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

रिकाम्या जागा भरा
Advertisements

उत्तर

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

Explanation:

Given `lim_(x -> 0) (sin mx cot  x/sqrt(3))` = 2

= `lim_((x -> 0),(because mx -> 0)) (sin mx)/(mx) xx mx  lim_(x -> 0) (cot  x/sqrt(3))` = 2

= `1 xx mx xx lim_(x -> 0) 1/(tan  /sqrt(3))` = 2

= `lim_(x -> 0) mx xx (x/sqrt(3))/(x/sqrt(3) * tan  x/sqrt(3))` = 2

= `(mx)/(x/sqrt(3)) (1)` = 2

⇒ `sqrt(3)m` = 2

⇒ m = `2/sqrt(3) = (2sqrt(3))/3`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 78 | पृष्ठ २४५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x -> 0) (cosec x -  cot x)`


Evaluate the following limit.

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following :

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


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


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


Evaluate `lim_(x -> pi/2) (secx - tanx)`


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


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


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


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


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


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


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


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


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)`


x cos x


`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`


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


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


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


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


If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` 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 -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.


Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = 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×