eˣexpands to 1+x+x²/2!+x³/3!+...
f(x)=eˣ⁺¹=e(eˣ)≈e(1+x) when x is very small (close to zero).
f(e(1+x))=eᵉ⁽¹⁺ˣ⁾⁺¹=eᵉ⁺ᵉˣ⁺¹=e(eᵉeᵉˣ).
f(e)=eᵉ⁺¹=e(eᵉ).
f(f(x))-f(e)=e(eᵉ)(eᵉˣ-1)=e(eᵉ)(1+ex-1)=e²xeᵉ.
(f(f(x))-f(e))/x=e²eᵉ=111.98 approx.
As x→0, the expression→e²eᵉ.