English

Evaluate the following limit : limx→0[1-cos(nx)1-cos(mx)]

Advertisements
Advertisements

Question

Evaluate the following limit :

`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`

Sum
Advertisements

Solution

`lim_(x -> 0)(1 - cos"n"x)/(1 - cos"m"x)`

= `lim_(x -> 0) (1 - cos"n"x)/(1 - cos"m"x) xx (1 + cos "n"x)/(1 + cos "n"x) xx (1 + cos"m"x)/(1 + cos"m"x)`

= `lim_(x -> 0) ((1 - cos^2"n"x)(1 + cos"m"x))/((1 - cos^2"m"x)(1 + cos "n"x))`

= `lim_(x -> 0) (sin^2"n"x(1 + cos "m"x))/(sin^2"m"x(1 + cos "n"x))`

= `lim_(x -> 0) (((sin^2"n"x)/("n"^2x^2))(1 + cos "m"x))/(((sin^2"m"x)/("m"^2x^2))(1 + cos "n"x)) xx "n"^2/"m"^2`  ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]

= `"n"^2/"m"^2 ([lim_(x -> 0) (sin"n"x)/("n"x)]^2 xx [lim_(x -> 0) (1 + cos "m"x)])/([lim_(x -> 0) (sin"m"x)/("m"x)]^2 xx [lim_(x -> 0) (1 + cos "n"x)])`

= `"n"^2/"m"^2 (1^2*(1 + cos 0))/(1^2*(1 + cos 0))  ...[because  x -> 0  therefore "m"x, "n"x -> 0 and lim_(theta -> 0)  sintheta/theta = 1]`

= `"n"^2/"m"^2`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Limits - Exercise 7.4 [Page 148]

RELATED QUESTIONS

Evaluate the following limit.

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


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 -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =


Evaluate the following :

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


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


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


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


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


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


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


Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`


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


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


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


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


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


cos (x2 + 1)


`x^(2/3)`


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


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


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


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


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


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


`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 ______.


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 ______.


The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin  x^2/4 log(1 + 3x))`, is ______.


`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×