English

Evaluate the following : limx→a[sinx-sinax-a] - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following :

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

Sum
Advertisements

Solution

`lim_(x -> "a") (sinx - sin"a")/(x - "a")`

Put x = a + h,

∴ x – a = h

As x → a, h → 0

∴ `lim_(x -> "a") (sinx - sin"a")/(x - "a")`

= `lim_("h" -> 0) (sin "a" + "h" - sin"a")/"h"`

= `lim_("h" -> 0) (2cos (("a" + "h" + "a")/2) sin(("a" + "h" - "a")/2))/"h"`

= `lim_("h" -> 0) (2cos("a" + "h"/2) sin  "h"/2)/"h"`

= `lim_("h" -> 0) cos ("a" + "h"/2) * lim_("h" -> 0) (sin("h"/2))/(("h"/2))`

= `cos ("a" + 0)(1)  ...[because "h" -> 0, "h"/2 -> 0  "and" lim_(theta -> 0) sintheta/theta = 1]`

= cos a

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Limits - Miscellaneous Exercise 7.2 [Page 159]

APPEARS IN

RELATED QUESTIONS

Evaluate the following limit.

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


Evaluate the following limit :

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


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]`


Evaluate the following :

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


`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______ 


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


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


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 -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`


Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`


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


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


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 -> pi/4)  (sin x - cosx)/(x - pi/4)`


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


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


`(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) sinx/(sqrt(x + 1) - sqrt(1 - 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)` = ______.


`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 ______.


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×