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

Evaluate the following limit : limx→0[5x+33-2x]2x - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`

बेरीज
Advertisements

उत्तर

`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`

= `lim_(x -> 0) [(3 + 5x)/(3 - 2x)]^(2/x)`

= `lim_(x -> 0) [(1 + (5x)/3)/(1 - (2x)/3)]^(2/x)`

= `lim_(x -> 0)((1 + (5x)/3)^(2/x))/((1 - (2x)/3)^(2/x))`

= `(lim_(x -> 0) (1 + (5x)/3)^(2/x))/(lim_(x -> 0) (1 - (2x)/3)^(2/x))`

= `(lim_(x -> 0)[(1 + (5x)/3)^(3/(5x))]^(10/3))/(lim_(x -> 0)[(1 - (2x)/3)^((-3)/(2x))]^((-4)/3)`

= `([lim_(x -> 0) (1 + (5x)/3)^(3/(5x))]^(10/3))/([lim_(x -> 0) (1 - (2x)/3)^((-3)/(2x))]^((-4)/3)`

= `("e"^(10/3))/("e"^((-4)/3))   ...[(because x -> 0  therefore (5x)/3 -> 0"," (-2x)/3 -> 0),(and lim_(x -> 0) (1 + x)^(1/x) = "e")]`

= `"e"^(14/3)`.

shaalaa.com
Limits of Exponential and Logarithmic Functions
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Exercise 7.6 [पृष्ठ १५४]

APPEARS IN

संबंधित प्रश्‍न

Evaluate the following: `lim_(x -> 0)[(log(3 - x) - log(3 + x))/x]`


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


Evaluate the following: `lim_(x -> 0)[(15^x - 5^x - 3^x +1)/x^2]`


Evaluate the following: `lim_(x -> 2) [(3^(x/2) - 3)/(3^x - 9)]`


Evaluate the following: `lim_(x -> 0)[((49)^x- 2(35)^x + (25)^x)/x^2]`


Evaluate the following Limits: `lim_(x -> 0)((1 - x)^5 - 1)/((1 - x)^3 - 1)`


Evaluate the following Limits: `lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/x]`


Evaluate the following Limits: `lim_(x -> 0) ("e"^x + e^(-x) - 2)/x^2`


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


Evaluate the following Limits: `lim_(x -> 0) [("a"^(4x) - 1)/("b"^(2x) - 1)]`


Evaluate the following Limits: `lim_(x -> 0)[(log(4 - x) - log(4 + x))/x]`


Evaluate the following limit : 

`lim_(x -> 0) [(6^x + 5^x + 4^x - 3^(x + 1))/sinx]`


Evaluate the following limit : 

`lim_(x -> 0) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`


Evaluate the following limit : 

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


Evaluate the following limit : 

`lim_(x -> 0) [(5 + 7x)/(5 - 3x)]^(1/(3x))`


Evaluate the following limit : 

`lim_(x ->0) [("a"^x - "b"^x)/(sin(4x) - sin(2x))]`


Evaluate the following limit :

`lim_(x -> 0) [((49)^x - 2(35)^x + (25)^x)/(sinx* log(1 + 2x))]`


Evaluate the following :

`lim_(x -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`


Evaluate the following :

`lim_(x -> 0) [("a"^(3x) - "a"^(2x) - "a"^x + 1)/(x*tanx)]`


Evaluate the following :

`lim_(x -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`


Evaluate the following : 

`lim_(x -> 0) [((5^x - 1)^2)/((2^x - 1)log(1 + x))]`


If the function

f(x) = `(("e"^"kx" - 1)tan "kx")/"4x"^2, x ne 0`

= 16 , x = 0

is continuous at x = 0, then k = ?


The value of `lim_{x→2} (e^{3x - 6} - 1)/(sin(2 - x))` is ______ 


`lim_(x -> 0) (15^x - 3^x - 5^x + 1)/(xtanx)` is equal to ______.


Evaluate the following Limit.

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


Evaluate the following limit :

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


Evaluate the following:

`lim_(x->0)[((25)^x - 2(5)^x + 1)/(x^2)]`


Evaluate the following:

`lim_(x -> 0)[((25)^x - 2(5)^x + 1)/x^2]`


Evaluate the limit: 

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


Evaluate the following:

`lim_(x->0)[((25)^x - 2(5)^x + 1)/x^2]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×