मराठी

Evaluate: limx→02-1+cosxsin2x - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

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

= `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x) xx (sqrt(2) + sqrt(1 + cosx))/(sqrt(2) + sqrt(1 + cos x))`

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

= `lim_(x -> 0) (1 - cos x)/(sin^2 xx [sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) (2 sin^2  x/2)/((2 sin  x/2 cos  x/2)^2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) (2 sin^2  x/2)/(4 sin^2  x/2 cos^2  x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) 2/(4 cos^2  x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

Taking limit, we get

= `2/(4 cos^2 0) xx 1/((sqrt(2) + sqrt(2))`

= `1/2 xx 1/(2sqrt(2))`

= `1/(4sqrt(2))`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 26 | पृष्ठ २४०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

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


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 -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


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


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


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


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


Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`


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


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


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


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


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


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


`x^(2/3)`


x cos 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/(sqrt(x + 1) - sqrt(1 - x)` is ______.


`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.


`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` 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) (sin mx cot  x/sqrt(3))` = 2, then m = ______. 


`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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×