English

Evaluate: limx→124x2-12x -1

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Given that `lim_(x -> 1/2) (4x^2 - 1)/(2x  - 1)`

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

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

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

Taking limit, we have

= `2 xx 1/2 + 1`

= 1 + 1

= 2.

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 239]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 2 | Page 239

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) cos x/(pi - x)`


Evaluate the following limit.

`lim_(x -> 0) (ax +  xcos x)/(b sin 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) [(1 - cos2theta)/theta^2]`


Evaluate the following limit :

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


Evaluate the following limit :

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


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 → π/3) ((tan^2x - 3)/(sec^3x - 8))` =


Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`


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)]`


Evaluate `lim_(x -> pi/2) (secx - tanx)`


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.


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


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 2x)/x^2`


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


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


`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * 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) (x^2 cosx)/(1 - cosx)` is ______.


`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.


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


`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.


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


`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×