English

Evaluate: limx→01+x3-1-x3x2

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Given that `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`

= `lim_(x -> 0) ([sqrt(1 + x^3) - sqrt(1 - x^3)][sqrt(1 + x^3) + sqrt(1 - x^3)])/(x^2[sqrt(1 + x^3) + sqrt(1 - x^2)])`

= `lim_(x -> 0) ((1 + x^3) - (1 - x^3))/(x^2[sqrt(1 + x^3) + sqrt(1 - x^3)])`

= `lim_(x -> 0) (1 + x^3 - 1 + x^3)/(x^2[sqrt(1 + x^3) + sqrt(1 - x^3)])`

= `lim_(x -> 0) (2x^3)/(x^2[sqrt(1 + x^3) + sqrt(1 - x^3)])`

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

= 0

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 240]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 11 | Page 240

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit.

`lim_(x -> (pi)/2) (tan 2x)/(x - pi/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 -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`


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)]`


Evaluate the following limit :

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


Evaluate the following :

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


Evaluate the following :

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


Evaluate the following :

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


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


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


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


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


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


Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`


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


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^(2/3)`


`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 -> pi) sinx/(x - pi)` is equal to ______.


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


`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.


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


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.


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


The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin  x^2/4 log(1 + 3x))`, is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×