English

Evaluate the following limit. limx→0cos2x-1cosx-1 - Mathematics

Advertisements
Advertisements

Question

Evaluate the following limit.

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

Sum
Advertisements

Solution

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

= `lim_(x → 0) (1 - 2 sin^2 x - 1)/(cosx - 1)`

= `lim_(x → 0) (2sin^2 x)/(1 - cosx)`

= `lim_(x → 0) (2(1 - cosx)(1 + cosx))/(1 - cosx)`

= `lim_(x → 0) 2(1 + cosx)`

= 2(1 + cos 0)

= 2 × 2 

= 4

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise 13.1 [Page 302]

APPEARS IN

NCERT Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise 13.1 | Q 17 | Page 302

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit.

`lim_(x ->0) cos x/(pi - x)`


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


Select the correct answer from the given alternatives.

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


Evaluate the following :

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


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


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


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


Find the derivative of f(x) = `sqrt(sinx)`, by first principle.


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


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


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


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


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


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


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


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


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


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


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


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


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 ______.


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


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


If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×