Advertisements
Advertisements
प्रश्न
Evaluate the following Limits: `lim_(x -> 0) ("e"^x + e^(-x) - 2)/x^2`
Advertisements
उत्तर
`lim_(x -> 0) ("e"^x + e^(-x) - 2)/x^2`
= `lim_(x -> 0) ("e"^x + 1/"e"^x - 2)/x^2`
= `lim_(x -> 0) (("e"^x)^2 + 1 - 2"e"^x)/(x^2*"e"^x`
= `lim_(x -> 0) ((e^x - 1)^2)/(x^2*"e"^x)`
= `lim_(x -> 0) [(("e"^x - 1)/x)^2 xx 1/"e"^x]`
= `lim_(x -> 0) (("e"^x - 1)/x)^2 xx 1/(lim_(x -> 0) "e"^x`
= `(1)^2 xx 1/"e"^0 ...[lim_(x -> 0) ("e"^x - 1)/x = 1]`
= `1 xx 1/1`
= 1
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0)[(9^x - 5^x)/(4^x - 1)]`
Evaluate the following: `lim_(x -> 0)[(5^x + 3^x - 2^x - 1)/x]`
Evaluate the following: `lim_(x -> 0) [(3^x + 3^-x - 2)/x^2]`
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: `lim_(x -> 2) [(3^(x/2) - 3)/(3^x - 9)]`
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 Limits: `lim_(x -> 0)[("a"^(3x) - "a"^(2x) - "a"^x + 1)/x^2]`
Evaluate the following limit :
`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`
Evaluate the following limit :
`lim_(x -> 0) [(5 + 7x)/(5 - 3x)]^(1/(3x))`
Evaluate the following :
`lim_(x -> 2) [(logx - log2)/(x - 2)]`
`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 limit:
`lim_(z->2)[(z^2-5x+6)/(z^2-4)]`
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]`
\[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}=\]
