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

Evaluate the following limit : limu→1[u4-1u3-1] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

`lim_(u -> 1) [(u^4 - 1)/(u^3 - 1)]`

बेरीज
Advertisements

उत्तर

`lim_(u -> 1) [(u^4 - 1)/(u^3 - 1)]`

= `lim_(u -> 1) ((u^4 - 1^4)/(u - 1))/((u^3 - 1^3)/(u - 1))   ...[(because  u  -> 1"," therefore u ≠ 1","),(therefore u - 1 ≠ 0)]`

= `(lim_(u -> 1) ((u^4 - 1^4)/(u - 1)))/(lim_(u -> 1)((u^3 - 1^3)/(u - 1))`

= `(4(1)^3)/(3(1)^2)   ...[because  lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `4/3`

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

APPEARS IN

संबंधित प्रश्‍न

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


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


Evaluate the following limits: `lim_(x -> 3) [1/(x - 3) - (9x)/(x^3 - 27)]`


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


Evaluate the following limits: `lim_(y -> 1/2)[(1 - 8y^3)/(y - 4y^3)]`


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


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


Evaluate the following limit:

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


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


Evaluate the following limit:

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limits

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


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->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)]`


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


Evaluate the following limits:

`lim_(z→2)[( z^2 - 5 z + 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->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×