Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$ 
Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then:
  $$
\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\text{(assume $|bx| < 2$)}
$$

Expanding $e^{ax}$ first, we get:
$$
1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}
$$
and expanding $(2+bx)^{-1}$ we get:
$$
\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}
$$
I tried dividing the expressions and comparing the coefficients, but didn't really get anywhere. Is there an easier way to do this?

Hint: Don't divide, Multiply!
$$\frac{e^{ax}}{2+bx}=e^{ax}(2+bx)^{-1}=\left(1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}+\cdots\right)\left(\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}\cdots\right)$$
Can you take it from here?
