Advertisements
Advertisements
प्रश्न
Evaluate the following: `lim_(x -> 0) [(3^x + 3^-x - 2)/x^2]`
Advertisements
उत्तर
`lim_(x -> 0) (3^x + 3^-x - 2)/x^2`
= `lim_(x -> 0) (3^x + 1/3^x - 2)/x^2`
= `lim_(x -> 0) ((3^x)^2 + 1 - 2(3^x))/(3^x*x^2)`
= `lim_(x -> 0) ((3^x - 1)^2)/(x^2*(3^x)` ...[∵ a2 – 2ab + b2 = (a – b)2]
= `lim_(x -> 0)((3^x - 1)/x)^2 xx 1/3^x`
= `lim_(x -> 0) ((3^x - 1)/x)^2 xx 1/(lim_(x -> 0) (3^x)`
= `(log3)^2 xx 1/3^0`
= `(log3)^2 xx 1/1 ....[lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= (log3)2
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0)[(log(2 + x) - log( 2 - x))/x]`
Evaluate the following: `lim_(x -> 0)[(log(3 - x) - log(3 + x))/x]`
Evaluate the following: `lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
Evaluate the following:
`lim_(x ->0)[((25)^x - 2(5)^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) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`
Evaluate the following limit :
`lim_(x -> 0) [(3^x + 3^-x - 2)/(x*tanx)]`
Evaluate the following limit :
`lim_(x -> 0) [((49)^x - 2(35)^x + (25)^x)/(sinx* log(1 + 2x))]`
Evaluate the following :
`lim_(x -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`
The value of `lim_{x→2} (e^{3x - 6} - 1)/(sin(2 - x))` is ______
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]`
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)]`
