Advertisements
Advertisements
Question
Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`
Advertisements
Solution
`lim_(x -> 0)(x(6^x - 3^x))/((2^x - 1)*log(1 + x))`
= `lim_(x -> 0)(x(3^x*2^x - 3^x))/((2^x - 1)*log(1 + x))`
= `lim_(x -> 0) (x*3^x(2^x - 1))/((2^x - 1)*log(1 + x)`
= `lim_(x -> 0) (x*3^x)/(log (1 + x)) ...[("As" x -> 0"," 2^x -> 2^0),("i.e." 2^x -> 1 therefore 2^x ≠ 1),(therefore 2^x - 1 ≠ 0)]`
= `lim_(x -> 0) (3^x)/((log(1 + x))/x`
= `(lim_(x -> 0) 3^x)/(lim_(x -> 0) (log(1 + x))/x`
= `3^0/1 ...[lim_(x -> 0) (log(1 + x))/x = 1]`
= 1
APPEARS IN
RELATED QUESTIONS
Evaluate the following: `lim_(x -> 0)[(9^x - 5^x)/(4^x - 1)]`
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)[(15^x - 5^x - 3^x +1)/x^2]`
Evaluate the following Limits: `lim_(x -> 0)[(log 100 + log (0.01 + 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) [(3 + x)/(3 - x)]^(1/x)`
Evaluate the following limit :
`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/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) ((3 + 5x)/(3 - 4x))^(1/x)` =
Select the correct answer from the given alternatives.
`lim_(x -> 0) [(x*log(1 + 3x))/("e"^(3x) - 1)^2]` =
Evaluate the following :
`lim_(x -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`
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) (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)]`
