Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 19 / b / Solution

Use the Maclaurin series on the positive side:In the last expression the original square root covers $e^x-1$, whose leading term is $x>0$. Hence $g/f\to 0$ and $f/h\to 0$. All three are positive for sufficiently small positive $x$, soThus $h$ is greatest and $g$ least in the stated limit.