Solution
= Solution
Use the <Maclaurin series> on the positive side:
$$
f(x)=\frac{x^2}{2}+O(x^4),\qquad
g(x)=\frac{x^3}{6}+O(x^5),\qquad
h(x)=\sqrt{x}\,[1+O(x)].
$$
In the last expression the original square root covers $e^x-1$, whose leading term is $x>0$. Hence $g/f\to0$ and $f/h\to0$. All three are positive for sufficiently small positive $x$, so
$$
\boxed{g(x)<f(x)<h(x)\quad\text{for sufficiently small }x>0.}
$$
Thus $h$ is greatest and $g$ least in the stated limit.