Advertisements
Advertisements
प्रश्न
Evaluate the following: `lim_(x -> 0)[(9^x - 5^x)/(4^x - 1)]`
Advertisements
उत्तर
`lim_(x -> 0)(9^x - 5^x)/(4^x - 1)`
= `lim_(x -> 0) (9^x - 1 + 1 - 5^x)/(4^x - 1)`
= `lim_(x -> 0) ((9^x - 1) - (5^x - 1))/(4^x - 1)`
= `lim_(x -> 0) ((9^x - 1 - 5^x - 1)/x)/((4^x - 1)/x` ...[∵ x → 0, ∴ x ≠ 0]
= `lim_(x -> 0) (((9^x - 1)/x) - ((5^x - 1)/x))/(((4^x - 1)/x)`
= `(lim_(x -> 0) (9^x - 1)/x - lim_(x -> 0) (5^x - 1)/x)/(lim_(x -> 0) (4^x - 1)/x)`
= `(log9 - log5)/(log4) ...[because lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= `1/(log4)log(9/5)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0) [(2^x - 1)^2/((3^x - 1) xx log (1 + x))]`
Evaluate the following:
`lim_(x ->0)[((25)^x - 2(5)^x + 1)/x^2]`
Evaluate the following Limits: `lim_(x -> 0)[(5^x - 1)/x]`
Evaluate the following Limits: `lim_(x -> 0)((1 - x)^5 - 1)/((1 - x)^3 - 1)`
Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`
Evaluate the following Limits: `lim_(x -> 0)[("a"^(3x) - "a"^(2x) - "a"^x + 1)/x^2]`
Evaluate the following Limits: `lim_(x -> 0) [("a"^(4x) - 1)/("b"^(2x) - 1)]`
Evaluate the following Limits: `lim_(x -> 0)[(log 100 + log (0.01 + x))/x]`
Evaluate the following limit :
`lim_(x -> 0) [(9^x - 5^x)/(4^x - 1)]`
Evaluate the following limit :
`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`
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^(sinx) - 1)^3/((3^x - 1).tan x.log(1 + x))]` =
Evaluate the following :
`lim_(x -> 2) [(logx - log2)/(x - 2)]`
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 = ?
`lim_(x -> 0) (sin^4 3x)/x^4` = ________.
Evaluate the following Limit.
`lim_(x->1)[(x^3-1)/(x^2+5x-6)]`
Evaluate the following :
`lim_(x -> 0) [((25)^x - 2(5)^x + 1)/x^2]`
