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

Evaluate the following limit : limx→0[8sinx-2tanxe2x-1]

Advertisements
Advertisements

प्रश्न

Evaluate the following limit : 

`lim_(x -> 0) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`

मूल्यांकन
Advertisements

उत्तर

`lim_(x -> 0) (8^sinx - 2^tanx)/("e"^(2x) - 1)`

= `lim_(x -> 0) ((8^sinx - 1)(2^tanx - 1))/("e"^(2x) - 1)`

= `lim_(x -> 0) (((8^sinx - 1) - (2^tanx - 1))/x)/(("e"^(2x) - 1)/x)   ...[("Divide numerator and"),("Denominator by"  x.),(because x -> 0 therefore x ≠ 0)]`

= `(lim_(x -> 0) ((8^sinx - 1)/x - (2^tanx - 1)/x))/(lim_(x -> 0) ("e"^(2x) - 1)/x)`

= `(lim_(x -> 0) ((8^sinx - 1)/sinx* sinx/x - (2^tanx - 1)/tanx * tanx/x))/(lim_(x -> 0) (e^(2x) - 1)/x)`

= `((lim_(x -> 0) (8^sinx - 1)/sinx)(lim_(x -> 0) sinx/x) - (lim_(x -> 0) (2^tanx - 1)/tanx)(lim_(x -> 0) tanx/x))/((lim_(x -> 0) ("e"^(2x) - 1)/(2x)) xx 2)`

= `(log8(1) - (log2)(1))/((1) xx 2)  ...[(because x -> 0","  2x -> 0","),(sin x -> 0"," tanx -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log"a")]`

= `(log  8/2)/2`

= `(log4)/2`

= `(log(2)^2)/2`

= `(2log2)/2`

= log 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)[(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 -> 2) [(3^(x/2) - 3)/(3^x - 9)]`


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) ("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)[(log(4 - x) - log(4 + x))/x]`


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


Evaluate the following limit : 

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


Evaluate the following limit : 

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


Evaluate the following limit : 

`lim_(x ->0) [("a"^x - "b"^x)/(sin(4x) - sin(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) [(log(5 + x) - log(5 - x))/sinx]` =


Select the correct answer from the given alternatives.

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


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


`lim_{x→∞} ((3x + 3)^40(9x - 3)^5)/(3x + 1)^45` = ______ 


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 ______ 


`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 ______ 


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->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]`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×