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

Evaluate the following limit : limx→0[5x+3x-2x-1x]

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

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

मूल्यांकन
Advertisements

उत्तर

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

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

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

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

= log 5 + log 3 – log 2    ...`[because  lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= `log((5 xx 3)/2)`

= `log(15/2)`

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)[(9^x - 5^x)/(4^x - 1)]`


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)[(5^x - 1)/x]`


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


Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`


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) [(9^x - 5^x)/(4^x - 1)]`


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


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) ((15^x - 3^x - 5^x + 1)/sin^2x)` =


Select the correct answer from the given alternatives.

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


Select the correct answer from the given alternatives.

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


Select the correct answer from the given alternatives.

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


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


The value of `lim_{x→-∞} (sqrt(5x^2 + 4x + 7))/(5x + 4)` is ______ 


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 ______.


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


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 limit: 

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×