<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] $$\eqalign{\ln(1+x^2) &= - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4}+ \ldots\cr
 \sqrt{1+x^4} - 1 &= \frac{x^4}{2} - \frac{x^8}{8} + \ldots\cr
 \frac{\ln(1+x^2)}{\sqrt{1+x^4}-1} &= \frac{x^4 \left(-\dfrac{1}{2} + \dfrac{x^2}{3} + \ldots\right)}{x^4 \left(\dfrac{1}{2} - \dfrac{x^4}{8} + \ldots\right)}\cr
&= -1 + \frac{2}{3} x^2 + \ldots}$$
</s>