Expansion of a function
Let $a>0$ be fixed. I would like to show that for $x>0$ we have $$
\frac{{1 - \left( {a/x} \right)^{\frac{4}{3}} }}{{1 - \left( {a/x}
\right)^2 }} = \frac{{1 - a^{\frac{4}{3}} }}{{1 - a^2 }} + \left( {x - 1}
\right)f\left( {x,a} \right) $$ with $\left| {f\left( {x,a} \right)}
\right| \le 1$. I got in trouble with Taylor's formula since the
derivative is not bounded as $x\to 0^+$. Any idea?