<s>[INST] 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?
 [/INST] Hint 
As Integrator said, multiply.
Another simple way  : consider the function $$e^{a x}=(2+b x)(\frac{1}{2}+\frac{x^2}{4}-cx^3)$$ Use the Taylor series for the lhs (just as you did) and expand the rhs. Now, identify as many terms as you can. You will then get the following equations $$ a-\frac{b}{2}=0$$ $$a^2-1=0$$ $$\frac{a^3}{6}-\frac{b}{4}+2 c=0$$
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