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

Evaluate the following limit : limx→0[3x+3-x-2x⋅tanx] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

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

बेरीज
Advertisements

उत्तर

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

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

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

= `lim_(x -> 0) (3^x - 1)^2/(3^x *xtanx)`

= `lim_(x -> 0) ((3^x - 1)/x)^2/(3^x * (tanx/x))`  ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]

= `(lim_(x -> 0) (3^x - 1)/x)^2/((lim_(x -> 0) 3^x) xx (lim_(x -> 0) tanx/x)`

= `(log3)^2/(3^0*1)    ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= (log 3)2

shaalaa.com
  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) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`


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


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


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


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) [((25)^x - 2(5)^x + 1)/(x*sinx)]`


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 -> 0) [(log(5 + x) - log(5 - x))/sinx]` =


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


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))]`


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


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


`lim_(x -> 0) (log(1 + (5x)/2))/x` is equal to ______.


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


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 ______ 


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


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 :

`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×