हिंदी

Evaluate the following limit : limx→0[3+x3-x]1x

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

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

योग
Advertisements

उत्तर

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

= `lim_(x -> 0) [(1 + x/3)/(1 - x/3)]^(1/x)   ...[("Divide numerator and"),("denominator by 3")]`

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

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

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

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

= `"e"^(1/3 + 1/3)`

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.6 [पृष्ठ १५४]

APPEARS IN

संबंधित प्रश्न

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


Evaluate the following: `lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`


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


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

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


Evaluate the following limit : 

`lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/sinx]`


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

`lim_(x -> 0)[(15^x - 5^x - 3^x + 1)/(x*sinx)]`


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 -> pi/2) ((3^(cosx) - 1)/(pi/2 - x))` =


Evaluate the following :

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


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→-∞} (sqrt(5x^2 + 4x + 7))/(5x + 4)` is ______ 


If f: R → R is defined by f(x) = [x - 2] + |x - 5| for x ∈ R, then `lim_{x→2^-} f(x)` is equal to ______ 


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


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


Evaluate the following Limit.

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


Evaluate the following :

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


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]`


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)]`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×