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]

संबंधित प्रश्‍न

Evaluate the following: `lim_(x -> 0)[(9^x - 5^x)/(4^x - 1)]`


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


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


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


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


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


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)[(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 100 + log (0.01 + x))/x]`


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


Evaluate the following limit : 

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


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))]`


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


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 ______


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->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 limit: 

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×