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

Evaluate the following limit : limx→0[log(3-x)-log(3+x)x]

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

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

बेरीज
Advertisements

उत्तर

`lim_(x -> 0) (log(3 - x) - log(3 + x))/x`

= `lim_(x -> 0) 1/x log ((3 - x)/(3 + x))`

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

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

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

= `log[{lim_(x -> 0)(1 - x/3)^((-3)/x)}^((-1)/3)/{lim_(x -> 0)(1 + x/3)^(3/x)}^(1/3)]`

= `log("e"^(-1/3)/("e"^(1/3)))   ...[because x -> 0","  ± x/3 -> 0  "and" lim_(x -> 0) (1 + x)^(1/x) = "e"]`

= `log "e"^(-(2)/3)`

= `-2/3*log "e"`

= `-2/3(1)`

= `-2/3`

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

संबंधित प्रश्‍न

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


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


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 Limits: `lim_(x -> 0)[(5^x - 1)/x]`


Evaluate the following Limits: `lim_(x -> 0)[(log(1 + 9x))/x]`


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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

`lim_(x -> 0)[(2^x - 1)^3/((3^x - 1)*sinx*log(1 + x))]`


Evaluate the following limit :

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


Select the correct answer from the given alternatives.

`lim_(x→0)[(3^(sinx) - 1)^3/((3^x - 1).tan x.log(1 + x))]` =


Select the correct answer from the given alternatives.

`lim_(x -> 3) [(5^(x - 3) - 4^(x - 3))/(sin(x - 3))]` =


Evaluate the following :

`lim_(x -> 2) [(logx - log2)/(x - 2)]`


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 = ?


lf the function f(x) satisfies `lim_{x→1}(2f(x) - 5)/(2(x^2 - 1)) = e`, then `lim_{x→1}f(x)` is ______ 


`lim_(x -> 0) (sin^4 3x)/x^4` = ________.


The value of `lim_{x→0} (1 + sinx - cosx + log_e(1 - x))/x^3` is ______


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 following:

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


Evaluate the following limit :

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


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 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 following:

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


\[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}=\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×