हिंदी

Evaluate the following : limx→0[(5x-1)2(2x-1)log(1+x)]

Advertisements
Advertisements

प्रश्न

Evaluate the following : 

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

योग
Advertisements

उत्तर

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

= `lim_(x -> 0)((5^x - 1)^2/x^2)/(((2^x - 1)*log(1 + x))/x^2)   ...[("Divide numerator and"),("denominator by"  x^2.),(because x -> 0","  x ≠ 0),(therefore x^2 ≠ 0)]`

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

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

= `(log5)^2/(log2 xx 1)  ....[lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= `(log5)^2/log2`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.2 | Q II. (14) | पृष्ठ १५९

संबंधित प्रश्न

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


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


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


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


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


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


Evaluate the following Limits: `lim_(x -> 0)[(log 100 + log (0.01 + x))/x]`


Evaluate the following limit : 

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


Evaluate the following limit : 

`lim_(x -> 0)[(4x + 1)/(1 - 4x)]^(1/x)`


Evaluate the following limit : 

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


Evaluate the following limit : 

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


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)]`


The value of `lim_{x→0}{(a^x + b^x + c^x + d^x)/4}^{1/x}` is ______ 


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) (log(1 + (5x)/2))/x` is equal to ______.


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


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

`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]`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×