मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Evaluate the following limit : limx→0(secx-1x2) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

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

बेरीज

उत्तर

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

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

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

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

= `lim_(x -> 0) [(tanx/x)^2 xx 1/(secx + 1)]`

= `lim_(x -> 0) (tanx/x)^2 xx lim_(x -> 0) 1/(secx + 1)`

= `(1)^2 xx 1/(1 + 1)`

= `1/2`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Exercise 7.4 [पृष्ठ १४८]

APPEARS IN

संबंधित प्रश्‍न

Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`


Evaluate the following limit :

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


Evaluate the following limit :

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


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


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


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


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


Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`


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


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


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))`


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


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


`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` 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 -> 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 ______.


If L = `lim_(x→∞)(x^2sin  1/x - x)/(1 - |x|)`, then value of L is ______.


If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.


`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×