English

Evaluate the following : limx→1[abx-axbx2-1] - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following :

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

Sum
Advertisements

Solution

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

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

= `lim_(x -> 1) ("ab"("b"^(x - 1) - "a"^(x - 1)))/((x - 1)(x + 1))`

Put x = 1 + h, ∴ x – 1 = h

As x → 1, h → 0

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

= `lim_("h" -> 0) ("ab"("b"^"h" - "a"^"h"))/("h"(1 + "h" + 1))`

= `"ab" lim_("h" -> 0) ("b"^"h" - 1 + 1 - "a"^"h")/("h"(2 + "h"))`

= `"ab" lim_("h" -> 0) (("b"^"h" - 1) - ("a"^"h" - 1))/("h"(2 + "h"))`

= `"ab" lim_("h" -> 0) 1/(2 + "h") (("b"^"h" - 1)/"h" - ("a"^"h" - 1)/"h")`

= `"ab"* 1/(lim_("h" -> 0)(2 + "h")) (lim_("h" -> 0) ("b"^"h" - 1)/"h" - lim_("h" -> 0) ("a"^"h" - 1)/"h")`

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

= `"ab"/2 log ("b"/"a")`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Limits - Miscellaneous Exercise 7.2 [Page 159]

APPEARS IN

RELATED QUESTIONS

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)[(15^x - 5^x - 3^x +1)/x^2]`


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)[("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 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)]^(1/x)`


Evaluate the following limit : 

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


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


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


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


Evaluate the following :

`lim_(x -> 2) [(logx - log2)/(x - 2)]`


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` = ______ 


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) (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 :

`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)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×