(i) `underset(x to oo)(Lt) e^(x) "sin" (Ke^(-x))`
(ii) `lim_(x to oo) (("sin"(3)/(4^(x)))/(tan (4)/(3^(x))))`
(iii) `lim_(x to oo) (1 - a^(4))^(-x) tan( b (1 - a^(4))^(x)) , -1 lt a lt 1, a != 0` and `b in R`

Explanation

We have,
`underset(x to oo)(lim) e^(x) sin (ke^(-x)) = underset(x to oo)(lim) (sin(ke^(-x)))/(ke^(-x)) K`
`= underset(e^(-x) to 0)(lim) (sin ke^(-x))/(ke^(-x)) k` `(x to oo implies e^(x) to oo implies e^(-x) = (1)/(e^(x)) to 0)`
`= 1 xx k = k`
(ii) `underset(x to oo)(lim) (sin(3)/(4^(x)))/(tan(4)/(3^(x))) = underset(x to oo)(lim) ((("sin" (3)/(4^(x))))/(((3)/(4^(x))))((3)/(4^(x))))/((("tan"(4)/(3^(x))))/(((4)/(3^(x))))((4)/(3^(x))))`,
Now `x to oo implies 4^(x) to oo, 3^(x) to oo`
`implies (3)/(4^(x)) to 0, (4)/(3^(x)) to 0`
`("sin"(3)/(4^(x))/((3)/(4^(x)) to 1 as x to 00, ("tan"(4)/(3^(x)))/((4)/(3^(x))) to 1 as x to oo` and `((3)/(4))^(x) to 0 as to oo`
Thus, `underset(x to oo)(lim) ("sin"(3)/(4^(x)))/("tan"(4)/(3^(x))) = 1 xx (3)/(4) underset(x to oo)(lim) ((3)/(4))^(x) = (3)/(4) 0 = 0`
We have,
`-1 lt a lt 1`
`implies - lt a^(4) lt 1`
`implies 0 lt - a^(4) lt 1`
`implies (1 - a^(4))^(x) to 0` as `x to oo`
Now `underset(x to oo)(lim) (1 - a^(4))^(-x) tan (b(1- a^(4))^(x))`
`underset(x to oo)(lim) (tan b(1 - a^(4))^(x))/(b(1 - a^(4))^(x))b`
`= 1 xx b = b`