<s>[INST] Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's Taylor series is close to this.
However, I'm not sure how to approach this. I tried to take several derivates of second term to see its value on $x_0 = 0$, but it becomes  complicated and I don't see general formula for $n$-th derivative at point zero to find $a$ and $b$.
 [/INST] Note you just need to compute the Taylor expansion of
\begin{align}
g(x)&=\cos x+bx^2\cos x-1-ax^2 \\[6px]
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}
+bx^2-\frac{bx^4}{2!}+\frac{bx^6}{4!}-1-ax^2+o(x^6)\\[6px]
&=\left(-\frac{1}{2}+b-a\right)x^2+
  \left(\frac{1}{24}-\frac{b}{2}\right)x^4+
  \left(-\frac{1}{6!}+\frac{b}{4!}\right)x^6+o(x^6)
\end{align}
Therefore
$$
\begin{cases}
a-b=-\dfrac{1}{2} \\[6px]
b=\dfrac{1}{12}
\end{cases}
$$
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