Originally Posted by

**Twig** I tried performing some hand calculations, and got the following result.

$\displaystyle f(x) = \frac{x^3}{3+5x} = x^3 \cdot \frac{1}{3+5x} $

Now I only performing a maclaurin series on the $\displaystyle \frac{1}{3+5x}$

term.

Let $\displaystyle h(x) \, \mbox{be the maclaurin series of }\frac{1}{3+5x}$

Then $\displaystyle f(x) \approx x^3 \cdot h(x) $

If I dont bother with the **restterm**, then we get something like (see picture).

And the two functions seem to be pretty close around $\displaystyle x=0$

You can see the maclaurin series in the picture...