Advertisements
Advertisements
Question
Evaluate the following Limits: `lim_(x -> 0)[("a"^(3x) - "a"^(2x) - "a"^x + 1)/x^2]`
Advertisements
Solution
`lim_(x -> 0)("a"^(3x) - "a"^(2x) - "a"^x + 1)/x^2`
= `lim_(x -> 0) ("a"^(2x)*"a"^x - "a"^(2x) - "a"^x + 1)/x^2`
= `lim_(x -> 0)("a"^(2x)("a"^x - 1) - 1("a"^x - 1))/x^2`
= `lim_(x -> 0) (("a"^x - 1)*("a"^(2x) - 1))/x^2`
= `lim_(x -> 0) (("a"^x - 1)/x xx("a"^(2x) - 1)/x)`
= `lim_(x -> 0)(("a"^x - 1)/x) xx lim_(x -> 0) (("a"^(2x) - 1)/(2x)) xx 2`
= `log"a"*(2log"a") ...[("As" x -> 0"," 2x -> 0 and),(lim_(x -> 0) ("a"^x - 1)/x = log "a")]`
= 2(log a)2
APPEARS IN
RELATED QUESTIONS
Evaluate the following: `lim_(x -> 0) [(3 + x)/(3 - x)]^(1/x)`
Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`
Evaluate the following limit :
`lim_(x -> 0) [(5^x + 3^x - 2^x - 1)/x]`
Evaluate the following limit :
`lim_(x -> 0) [(3 + x)/(3 - x)]^(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)]`
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)[(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 -> 0) [("a"^(3x) - "a"^(2x) - "a"^x + 1)/(x*tanx)]`
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) (log(1 + (5x)/2))/x` is equal to ______.
`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 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)]`
Evaluate the following:
`lim_(x->0) [((25)^x - 2(5)^x + 1)/x^2]`
\[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}=\]
