Advertisements
Advertisements
प्रश्न
Evaluate the following: `lim_(x -> 0)[(15^x - 5^x - 3^x +1)/x^2]`
Advertisements
उत्तर
`lim_(x -> 0)[(15^x - 5^x - 3^x +1)/x^2]`
= `lim_(x -> 0) (5^x*3^x - 5^x - 3^x + 1)/x^2`
= `lim_(x -> 0) (5^x (3^x- 1) - 1(3^x - 1))/x^2`
= `lim_(x -> 0) ((3^x - 1) (5^x - 1))/x^2`
= `lim_(x -> 0) ((3^x - 1)/x xx (5^x - 1)/x)`
= `lim_(x -> 0) (3^x - 1)/x xx lim_(x -> 0)(5^x - 1)/x`
= `log 3* log 5 ...[lim_(x -> 0) ("a"^x - 1)/x = log"a"]`
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0)[(log(2 + x) - log( 2 - x))/x]`
Evaluate the following: `lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
Evaluate the following: `lim_(x -> 0)[((49)^x- 2(35)^x + (25)^x)/x^2]`
Evaluate the following Limits: `lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/x]`
Evaluate the following limit :
`lim_(x -> 0) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`
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) [(log(3 - x) - log(3 + x))/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)]`
Evaluate the following limit :
`lim_(x -> 0) [((49)^x - 2(35)^x + (25)^x)/(sinx* log(1 + 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))]` =
If the function
f(x) = `(("e"^"kx" - 1)tan "kx")/"4x"^2, x ne 0`
= 16 , x = 0
is continuous at x = 0, then k = ?
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]`
