English

Evaluate limx→0 sin(2+x)-sin(2-x)x

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

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

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

= `lim_(x -> 0) (2cos 2sinx)/x`

= `2cos 2 lim_(x -> 0)  sinx/x`

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Solved Examples [Page 229]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 13 Limits and Derivatives
Solved Examples | Q 5 | Page 229

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit.

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


Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x -> 0) (ax +  xcos x)/(b sin x)`


Evaluate the following limit.

`lim_(x -> 0) (cosec x -  cot x)`


Evaluate the following limit :

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


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 → π/3) ((tan^2x - 3)/(sec^3x - 8))` =


Evaluate the following :

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


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`


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


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


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


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


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


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


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


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


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


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


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 cos x


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×