English

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

Advertisements
Advertisements

Question

Evaluate the following limit :

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

Sum
Advertisements

Solution

`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
  Is there an error in this question or solution?
Chapter 7: Limits - Exercise 7.4 [Page 148]

RELATED QUESTIONS

Evaluate the following limit.

`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`


Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`


Evaluate the following limit.

`lim_(x -> 0) (cosec x -  cot x)`


Evaluate the following limit :

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


Evaluate the following limit :

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


Select the correct answer from the given alternatives.

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


Evaluate the following :

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


Evaluate the following :

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


Evaluate the following :

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


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


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


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


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


Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`


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


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/4)  (sin x - cosx)/(x - pi/4)`


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


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


Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`


`x^(2/3)`


`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`


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


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


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


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


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


If `lim_(n→∞)sum_(k = 2)^ncos^-1(1 + sqrt((k - 1)(k + 2)(k + 1)k)/(k(k + 1))) = π/λ`, then the value of λ 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×