English

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

Advertisements
Advertisements

Question

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

Options

  • 3

  • 1

  • 0

  • 2

MCQ
Fill in the Blanks
Advertisements

Solution

`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
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 243]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 61 | Page 243

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit.

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


Evaluate the following limit.

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


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)]`


Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`


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


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


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


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


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


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


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


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


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


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


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


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


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


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


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


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


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


`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) (tan 2x - x)/(3x - sin x)` is equal to ______.


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


Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.


`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×