मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Evaluate the following limit : limx→-2[-2x-4x3+2x2] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

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

बेरीज
Advertisements

उत्तर

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

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

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

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

= `((-2))/(-2)^2`

= `(-2)/4`

= `(-1)/2`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Exercise 7.2 [पृष्ठ १४१]

संबंधित प्रश्‍न

Evaluate the following limits: `lim_(z -> 2) [(z^2 - 5z + 6)/(z^2 - 4)]`


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


Evaluate the following limits: `lim_(u -> 1)[(u^4 - 1)/(u^3 - 1)]`


Evaluate the following limits: `lim_(x -> 2)[(x^3 - 4x^2 + 4x)/(x^2 - 1)]`


Evaluate the following limit:

`lim_(x -> - 2)[(x^7 + x^5 + 160)/(x^3 + 8)]`


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


Evaluate the following Limits: `lim_(x -> 2)[((x - 2))/(2x^2 - 7x + 6)]`


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


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


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> 3) [(x^2 + 2x - 15)/(x^2 - 5x + 6)]`


Evaluate the following limit :

`lim_(x -> 2)[(x^3 - 4x^2 + 4x)/(x^2 - 1)]`


Evaluate the following limit :

`lim_(x -> 2) [(x^3 - 7x + 6)/(x^3 - 7x^2 + 16x - 12)]`


Evaluate the following limit :

`lim_(y -> 1/2) [(1 - 8y^3)/(y - 4y^3)]`


Evaluate the following limit :

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


Select the correct answer from the given alternatives.

`lim_(x -> 2) ((x^4 - 16)/(x^2 - 5x + 6))` =


Select the correct answer from the given alternatives.

`lim_(x -> -2)((x^7 + 128)/(x^3 + 8))` =


Select the correct answer from the given alternatives.

`lim_(x -> 5) ((sqrt(x + 4) - 3)/(sqrt(3x - 11) - 2))` =


Evaluate the following limit :

`lim_(x->-2)[(x^7 + x^5 +160)/(x^3+8)]`


Evaluate the following limit :

`lim_("x" -> -2) [("x"^7 + "x"^5 + 160)/("x"^3 +8)]`


Evaluate the following Limit.

`lim_(x->1)[(x^3 -1)/(x^2 +5x -6)]`


Evaluate the following Limit.

`lim_(x->1)[(x^3 -1)/(x^2 +5x -6)]`


Evaluate the following limit:

`lim_(z->2)[(z^2-5z+6)/(z^2-4)]`


Evaluate the following limit:

`lim_(z->2)[(z^2-5z+6)/(z^2-4)]`


Evaluate the following limit:

`lim_(x->-2) [(x^7 + x^5 +160)/(x^3 + 8)]`


Evaluate the following limit:

`lim_(x->-2) [(x^7 + x^5 +160)/(x^3 + 8)]`


Evaluate the following Limit.

`lim_(x->1)[(x^3-1)/(x^2+5x-6)]`


Evaluate the following limit:

`lim_(x->2) [(z^2 - 5_z + 6)/ (z^2 - 4)]` 


Evaluate the following Limit:

`lim_(x->1)[(x^3-1)/(x^2+5x-6)]`


Evaluate the following limit:

`lim_(x->-2)[(x^7 + x^5 + 160)/(x^3 + 8)]`


Evaluate the following limit:

`lim_(x ->1)[(x^3 - 1)/(x^2 + 5x - 6)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×