हिंदी

Evaluate the following limit : limx→2[1+2+x-3x-2] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

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

योग

उत्तर

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

`lim_(x -> 2) (sqrt(1 + sqrt(2 + x)) - sqrt(3))/(x - 2) xx (sqrt(1 + sqrt(2 + x)) + sqrt(3))/(sqrt(1 + sqrt(2 + x)) + sqrt(3))`

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

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

= `lim_(x -> 2) ((2 + x) - 4)/((x - 2)(sqrt(1 + sqrt(2 + x)) + sqrt(3))(sqrt(2 + x) + 2))`

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

= `lim_(x -> 2) 1/((sqrt(1 + sqrt(2 + x)) + sqrt(3))(sqrt(2 + x) + 2))   ...[(because  x -> 2","  x ≠ 2),(therefore x - 2 ≠ 0)]`

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

= `1/((sqrt(1 + sqrt(4)) + sqrt(3))(sqrt(2 + 2) + 2)`

= `1/((2sqrt(3))(4))`

= `1/(8sqrt(3))`

shaalaa.com
Rationalization Method
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.3 [पृष्ठ १४३]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 7 Limits
Exercise 7.3 | Q II. (3) | पृष्ठ १४३

संबंधित प्रश्न

Evaluate the following limits: `lim_(x -> 0) [(sqrt(6 + x + x^2) - sqrt(6))/x]`


Evaluate the following limits: `lim_(y -> 0) [(sqrt(1 - y^2) - sqrt(1 + y^2))/y^2]`


Evaluate the following limits: `lim_(x -> 2)[(x^2 - 4)/(sqrt(x + 2) - sqrt(3x - 2))]`


Evaluate the following limits: `lim_(x -> 1) [(x^2 + xsqrt(x) - 2)/(x - 1)]`


Evaluate the following limits:  `lim_(x -> 0)[(sqrt(1 + x^2) - sqrt(1 + x))/(sqrt(1 + x^3) - sqrt(1 + x))]`


``Evaluate the following limits: `lim_(x -> 4) [(x^2 + x - 20)/(sqrt(3x + 4) - 4)]`


Evaluate the following limits: `lim_(x -> 2) [(x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))]`


Evaluate the following limits: `lim_(y -> 2) [(2 - y)/(sqrt(3 - y) - 1)]`


Evaluate the following limits: `lim_(z -> 4) [(3 - sqrt(5 + z))/(1 - sqrt(5 - z))]`


Evaluate the following limit:

`lim_(x -> 0)[(sqrt(6 + x + x^2) - sqrt(6))/x]`


Evaluate the following limit :

`lim_(x -> 3)[(sqrt(2x + 3) - sqrt(4x - 3))/(x^2 - 9)]`


Evaluate the following limit :

`lim_(y -> 0)[(sqrt(1 - y^2) - sqrt(1 + y^2))/y^2]`


Evaluate the following limit :

`lim_(x -> 2) [(sqrt(2 + x) - sqrt(6 - x))/(sqrt(x) - sqrt(2))]`


Evaluate the following limit :

`lim_(x -> "a") [(sqrt("a" + 2x) - sqrt(3x))/(sqrt(3"a" + x) - 2sqrt(x))]`


Evaluate the following limit :

`lim_(x -> 2) [(x^2 - 4)/(sqrt(x + 2) - sqrt(3x - 2))]`


Evaluate the following limit :

`lim_(x -> 0)[(sqrt(x^2 + 9) - sqrt(2x^2 + 9))/(sqrt(3x^2 + 4) - sqrt(2x^2 + 4))]`


Evaluate the following limit :

`lim_(x -> 1) [(x^2 + xsqrt(x) - 2)/(x - 1)]`


Evaluate the Following limit :

`lim_(x -> 4) [(x^2 + x - 20)/(sqrt(3x + 4) - 4)]`


Evaluate the Following limit :

`lim_(z -> 4) [(3 - sqrt(5 + z))/(1 - sqrt(5 - z))]`


Evaluate the Following limit :

`lim_(x -> 0)[3/(xsqrt(9  - x)) - 1/x]`


Evaluate the following :

`lim_(x -> 0)[x]` ([*] is a greatest integer function.)


Evaluate the following :

If f(r) = πr2 then find `lim_("h" -> 0) [("f"("r" + "h") - "f"("r"))/"h"]`


Evaluate the following limit:

`lim_(x->0)[(sqrt(6 + x + x^2) - sqrt6)/x]`


Evaluate the following limit:

`lim_(x->0)[(sqrt(6+x+x^2)-sqrt6)/x]`


Evaluate the following limit:

`lim_(x->0)[(sqrt(6+x+x^2)-sqrt6)/x]`


Evaluate the following limit.

`lim_(x→0) [[sqrt(6 + x + x^2)- sqrt6]/x]`


Evaluate the following limit:

`lim_(x -> 0)[(sqrt(6 + x + x^2) - sqrt6)/x]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×