<s>[INST] Maclaurin series with zero denominator when evaluating derivative I have following function:
$\displaystyle f(x) = \frac{\ln(1+x^2) - x^2}{\sqrt{1+x^4} - 1}$
As you can see, when doing the quotient rule for the denominator in your head, the derivative of this function results to $0$ in the denominator when evaluating for $x = 0$. My question is, how is it possible that there exists a Maclaurin series according to Wolfram Alpha?
Do I overlook something?
 [/INST] Technically speaking, even the function itself is undefined at zero, as the denominator of $f(x)$ turns into zero when we try to plug in $x=0$. But this is a removable discontinuity, as the graph of $f(x)$ (available on the page that you linked to) shows. And moreover, with this discontinuity "removed" (or patched, so to speak), the function becomes smooth and analytic.
To understand better what's going on here, let's set up the Maclaurin expansions of the numerator and denominator separately:
$$\ln(1+x^2)-x^2=\left(x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots\right)-x^2=-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots;\\
\sqrt{1+x^4}-1=\left(1+\frac{x^4}{2}-\frac{x^8}{8}+\cdots\right)-1=\frac{x^4}{2}-\frac{x^8}{8}+\cdots.$$
You can see that both expansions start with $x^4$, so when we put them into a fraction, this $x^4$ can be factored out in both of them and canceled out:
$$\frac{\ln(1+x^2)-x^2}{\sqrt{1+x^4}-1}=\frac{-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots}{\frac{x^4}{2}-\frac{x^8}{8}+\cdots}=\frac{x^4\left(-\frac{1}{2}+\frac{x^2}{3}-\frac{x^4}{4}+\cdots\right)}{x^4\left(\frac{1}{2}-\frac{x^4}{8}+\cdots\right)}=\frac{-\frac{1}{2}+\frac{x^2}{3}-\frac{x^4}{4}+\cdots}{\frac{1}{2}-\frac{x^4}{8}+\cdots}=-1+\cdots,$$
see WolframAlpha for the rest. :-)
</s>