English

Evaluate: limx→asinx-sinax-a

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Given that `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`

= `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a)) xx (sqrt(x) + sqrt(a))/(sqrt(x) + sqrt(a))`

= `lim_(x -> a) ((sin x - sin a)(sqrt(x) + sqrt(a)))/(x - a)`

= `lim_(x -> a) ((2 cos  (x + a)/2 * sin  (x - a)/2)(sqrt(x) + sqrt(a)))/(x - a)`

= `lim_((x -> a),(because  (x - a)/2 -> 0)) (2 cos  (x + a)/2 * (sin  (x - a)/2)/(2 xx (x - a)/2)) (sqrt(x) + sqrt(a))`

= `lim_(x -> a) cos((x + a)/2)(sqrt(x) + sqrt(1))`  .....`[because  lim_((x - a)/2 -> 0)  (sin  (x - a)/2)/((x - a)/2) = 1]`

Taking limit we have

= `cos ((a + a)/2)(sqrt(a) + sqrt(a))`

= `cos a xx 2sqrt(a)`

= `2sqrt(a) * cos a`

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 240]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 24 | Page 240

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit :

`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


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)]`


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


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)]`


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


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


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


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


Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`


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


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


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


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


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


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


x cos x


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


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


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


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


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` 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 ______.


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×