मराठी

Limx→π4sec2x-2tanx-1 is equal to ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • 3

  • 1

  • 0

  • 2

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

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

Explanation:

Given, `lim_(x -> pi/4) (sec^2 x - 2)/(tan x - 1)`

= `lim_(x -> pi/4) (1 + tan^2 x - 2)/(tan x - 1)`

= `lim_(x -> pi/4) (tan^2x - 1)/(tanx - 1)`

= `lim_(x -> pi/4) ((tan x + 1)(tan x - 1))/((tan x - 1))`

= `lim_(x -> pi/4) (tan x + 1)`

= `tan  pi/4 + 1`

= 1 + 1

= 2

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 61 | पृष्ठ २४३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x → 0) x sec x`


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


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


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


Find the derivative of f(x) = `sqrt(sinx)`, by first principle.


`lim_(x -> 0) |x|/x` is equal to ______.


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


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


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


Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`


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


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


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


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


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


cos (x2 + 1)


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


x cos x


`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`


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


`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` 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) (tan 2x - x)/(3x - sin x)` is equal to ______.


`lim_(x -> 3^+) x/([x])` = ______.


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×