Advertisements
Advertisements
प्रश्न
Evaluate the following: `lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
Advertisements
उत्तर
`lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
= `lim_(x -> 0) ("a"^(3x) - 1 - "b"^(2x) - 1)/(log 1 + 4x)`
= `lim_(x -> 0) [("a"^(3x) - 1 - "b"^(2x) - 1)/x)/((log 1 + 4x)/x)`
= `(lim_(x -> 0) [("a"^(3x) - 1)/x - ("b"^(2x) - 1)/x])/(lim_(x -> 0) (log 1 + 4x)/(x)`
= `(lim_(x -> 0)[("a"^(3x) - 1)/(3x)] xx 3 - lim_(x -> 0)[("b"^(2x) - 1)/(2x)] xx 2)/(lim_(x -> 0) (log 1 + 4x)/(4x) xx 4)`
= `(3log "a" - 2 log"b")/(1 xx 4) ...[(because x -> 0"," 2x -> 0"," 3x -> 0),(4x -> 0 and lim_(x -> 0) ("a"^x - 1)/x = log "a"),(and lim_(x -> 0) (log (1 + x))/x = 1)]`
= `1/4(log"a"^3 - log"b"^2)`
= `1/4 log("a"^3/"b"^2)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0)[(9^x - 5^x)/(4^x - 1)]`
Evaluate the following Limits: `lim_(x -> 0)(1 + x/5)^(1/x)`
Evaluate the following Limits: `lim_(x -> 0)[((5^x - 1)^2)/(x*log(1 + x))]`
Evaluate the following Limits: `lim_(x -> 0)[(log(4 - x) - log(4 + x))/x]`
Evaluate the following limit :
`lim_(x -> 0) [(5^x + 3^x - 2^x - 1)/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) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((3 + 5x)/(3 - 4x))^(1/x)` =
Evaluate the following :
`lim_(x -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`
Evaluate the following :
`lim_(x -> 0) [((5^x - 1)^2)/((2^x - 1)log(1 + x))]`
`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) (log(1 + (5x)/2))/x` 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]`
