Advertisements
Advertisements
प्रश्न
Evaluate the following: `lim_(x -> 0)[(15^x - 5^x - 3^x +1)/x^2]`
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) [(3^x + 3^-x - 2)/x^2]`
Evaluate the following Limits: `lim_(x -> 0) ("e"^x + e^(-x) - 2)/x^2`
Evaluate the following Limits: `lim_(x -> 0)[(log(4 - x) - log(4 + x))/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) [(5 + 7x)/(5 - 3x)]^(1/(3x))`
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 -> 0) [(x*log(1 + 3x))/("e"^(3x) - 1)^2]` =
`lim_{x→∞} ((3x + 3)^40(9x - 3)^5)/(3x + 1)^45` = ______
The value of `lim_{x→-∞} (sqrt(5x^2 + 4x + 7))/(5x + 4)` is ______
`lim_(x -> 0) (15^x - 3^x - 5^x + 1)/(xtanx)` is equal to ______.
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]`
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)]`
