Solution: The focus of ellipse x2/5a+y2/(4a2+1) =1is on the X axis, so 5a >: 4a2+1> 0 = & gt4a 2-5a+ 1 & lt； 0 = & gt(a- 1)(4a- 1)& lt； 0 = & gt 1/4 & lt； A< 1 = > eccentricity e = c/a = √ (5a-4a2-1)/√ (5a) = (√ 5/5) √ (-4a-1a+5), because a ∈. a & gt0 = & gt4a+ 1/a≥2 √[ 4a *( 1/a)]= 4 = & gt； (4a + 1/a)∈[4，5]= >-4a- 1/a =-(4a+ 1/a)∈(-5，-4)= & gt； (-4a - 1/a + 5)∈(0， 1)= & gt； √(-4a - 1/a + 5)∈(0， 1)= & gt； e = (√5/5) √(-4a - 1/a + 5)∈(0，√5/5)。