Advertisements
Advertisements
प्रश्न
Evaluate the following:
`lim_(x ->0)[((25)^x - 2(5)^x + 1)/x^2]`
Advertisements
उत्तर
`lim_(x ->0)[((25)^x - 2(5)^x + 1)/x^2]`
= `lim_(x -> 0)[((5)^(2x) - 2(5)^x + 1)/x^2]`
= `lim_(x -> 0)[((5^x)^(2) - 2(5)^x + 1)/x^2]`
= `lim_(x -> 0) [(5^x - 1)^2/x^2]`
= `lim_(x -> 0) ((5^x - 1)/x)^2`
= `log5^2 ...[lim_(x -> 0) ("a"^x - 1)/x = log"a"]`
APPEARS IN
संबंधित प्रश्न
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) [(3^x + 3^-x - 2)/x^2]`
Evaluate the following Limits: `lim_(x -> 0)[(log(1 + 9x))/x]`
Evaluate the following Limits: `lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/x]`
Evaluate the following limit :
`lim_(x -> 0) [(9^x - 5^x)/(4^x - 1)]`
Evaluate the following limit :
`lim_(x -> 0) [(5^x + 3^x - 2^x - 1)/x]`
Evaluate the following limit :
`lim_(x -> 0) [(6^x + 5^x + 4^x - 3^(x + 1))/sinx]`
Evaluate the following limit :
`lim_(x -> 0) [(3 + x)/(3 - x)]^(1/x)`
Select the correct answer from the given alternatives.
`lim_(x -> 0) [(log(5 + x) - log(5 - x))/sinx]` =
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:
`lim_(x->0)[((25)^x -2(5)^x+1)/x^2]`
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]`
Evaluate the following limit :
`lim_(x->0)[(sqrt(6+x+x^2)-sqrt6)/x]`
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]`
