हिंदी

Evaluate the following : limx→a[xcosa-acosxx-a] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following :

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

योग
Advertisements

उत्तर

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

= `lim_(x -> "a") [(x cos "a" - "a" cos "a" + "a" cos "a" - "a" cos x)/(x - "a")]`   ...[Note this step]

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

= `lim_(x -> "a") [((x - "a") cos "a" + 2"a" sin (("a" + x)/2)((x - "a")/2))/(x - "a")]`

= `lim_(x -> "a") [((x - "a")cos"a")/(x - "a") + (2"a" sin (("a" + x)/2) sin((x - "a")/2))/(x - "a")]`

= `lim_(x -> "a") [cos"a" + "a" sin (("a" + x)/2)* (sin((x - "a")/2))/(((x - "a")/2))]`  ...[∵ x → a, x ≠ a, ∴ x – a ≠ 0]

= `lim_(x -> "a") cos"a" + "a"[lim_(x -> "a") sin(("a" + x)/2)] xx [lim_(x -> "a") sin((x - "a")/2)/((x - "a")/2)]` 

= `cos "a" + "a" sin (("a" + "a")/2) xx 1     ...[(because x -> "a" ","  x ≠ "a"  therefore (x - "a")/2 -> 0),(and lim_(theta -> 0) sintheta/theta = 1)]`

= cos a + a sin a

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.2 | Q II. (16) | पृष्ठ १५९

संबंधित प्रश्न

Evaluate the following limit.

`lim_(x -> 0) (ax +  xcos x)/(b sin 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 -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`


Evaluate the following limit :

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


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


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


Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`


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


Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.


Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`


If f(x) = x sinx, then f" `pi/2` is equal to ______.


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


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/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


cos (x2 + 1)


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


`x^(2/3)`


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


`lim_(x -> pi) sinx/(x - pi)` is equal to ______.


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


`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` 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 -> 0) (sin mx cot  x/sqrt(3))` = 2, then m = ______. 


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


`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×