Advertisements
Advertisements
प्रश्न
Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`
उत्तर
`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
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0)[(5^x + 3^x - 2^x - 1)/x]`
Evaluate the following: `lim_(x -> 0)[(log(2 + x) - log( 2 - x))/x]`
Evaluate the following: `lim_(x -> 0) [(2^x - 1)^2/((3^x - 1) xx log (1 + x))]`
Evaluate the following: `lim_(x -> 2) [(3^(x/2) - 3)/(3^x - 9)]`
Evaluate the following Limits: `lim_(x -> 0)[(log(1 + 9x))/x]`
Evaluate the following limit :
`lim_(x -> 0) [(9^x - 5^x)/(4^x - 1)]`
Evaluate the following limit :
`lim_(x -> 0) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`
Evaluate the following limit :
`lim_(x -> 0) [(5 + 7x)/(5 - 3x)]^(1/(3x))`
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) [((25)^x - 2(5)^x + 1)/(x*sinx)]`
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 -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`
Evaluate the following :
`lim_(x -> 0) [((5^x - 1)^2)/((2^x - 1)log(1 + x))]`
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 = ?
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 ______
Evaluate the following limit :
`lim(x>2)[(z^2 -5z+6)/(z^2-4)]`
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)]`