Problem with approximating function I want to approximate the expression $$ \frac{a z +1/2 b z^2 + c z^3}{a + b z + d z^2}.$$
This should be approximately equal to $z - \frac{b}{2a}z^2.$ But no matter the approach, I do not get this answer. Anyone who sees what I should have done?
Edit: some users had a fair point, in this case $z$ is very small

$$\begin{array}{rcl}\dfrac{az + \dfrac{1}{2} bz^2 + cz^3}{a + bz + dz^2} &=& \dfrac{az + bz^2 +dz^3-\dfrac{1}{2}bz^2-(d-c)z^3}{a +bz+dz^2}\\
&=&\dfrac{az + bz^2 +dz^3}{a+bz+dz^2}-\dfrac{\dfrac{1}{2}bz^2 +(d-c)z^3}{a+bz+dz^2}\\
&=&z -\dfrac{\dfrac{1}2b\left(z^2 +\dfrac{2(d-c)}{b}z^3\right)}{a+bz+dz^2}\\
&=& z - \dfrac{\dfrac{b}{2a}\left(az^2 +\dfrac{2a(d-c)}{b}z^3\right)}{a+bz+dz^2}\\
&=& z - \dfrac{b}{2a}\cdot \dfrac{az^2+bz^3+dz^4-dz^4+\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\
&=&z-\dfrac{b}{2a}z^2 +\dfrac{b}{2a}\cdot\dfrac{dz^4-\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\
\end{array}$$
This should be approximately equal to $z-\dfrac{b}{2a}z^2$ when $z$ is sufficiently small.
