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

Select the correct answer from the given alternatives. limx→3[5x-3-4x-3sin(x-3)] =

Advertisements
Advertisements

प्रश्न

Select the correct answer from the given alternatives.

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

पर्याय

  • log 5 – 4

  • `log  5/4`

  • `log5/log4`

  • `log5/4`

MCQ
Advertisements

उत्तर

`log  5/4`

Explanation;

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

Put x – 3 = h

∴ x = 3 + h

As → 3, h → 0

∴ Required limit

= `lim_("h" -> 0) (5^"h" - 4^"h")/(sin "h")`

= `lim_("h" -> 0) ((5^"h" - 1) - (4^"h" - 1))/sin"h"`

= `lim_("h" -> 0) (((5^"h" - 1))/"h" - ((4^"h" - 1))/"h")/(lim_("h" -> 0) sin"h"/"h"`

= `(log5 - log4)/1  ...[lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= `log(5/4)`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Miscellaneous Exercise 7.1 [पृष्ठ १५८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 7 Limits
Miscellaneous Exercise 7.1 | Q I. (14) | पृष्ठ १५८

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)[(log(2 + x) - log( 2 - x))/x]`


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


Evaluate the following: `lim_(x -> 2) [(3^(x/2) - 3)/(3^x - 9)]`


Evaluate the following:

`lim_(x ->0)[((25)^x - 2(5)^x + 1)/x^2]`


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


Evaluate the following Limits: `lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/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) [(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) [(5 + 7x)/(5 - 3x)]^(1/(3x))`


Evaluate the following limit : 

`lim_(x -> 0)[(15^x - 5^x - 3^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) [(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 -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`


Evaluate the following :

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


Evaluate the following :

`lim_(x -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`


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 ______.


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->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)]`


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×