If L[f(t)] = 2(s + 2), then ƒ(0+) and ƒ(∞)

1. 0, 2 respectively
2. 2, 0 respectively
3. 0, 1 respectively
4. 2/5, 0 respectively

2(s + 1)	=	2(s + 1)	= 2		s + 1
(s² + 2s + 5)		(s² + 1)² + 5			(s + 1)² + 4

This is standard Laplace Transform for the function

L_\| exp (– at) cos ωt _\| =	(s + a)
	(s + a)² + ω²

where a = 1, ω = 2
Hence ƒ(t) = 2e^{– t} cos 2t
ƒ(0) = ƒ(0+) = 2e^{– 0} cos 2 × 0 = 2
ƒ(∞) = 2e^{– ∞} t = 0

If L[f(t)] =	2(s + 2)	, then ƒ(0⁺) and ƒ(∞) are given by
	s² + 2s + 5