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 $b$.

Your function is 
$$\cos(x)-\frac{a}{b}-\frac{1-\frac{a}{b} }{  bx^2+1  }$$
$$=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{a}{b}-1+\frac{a}{b}+bx^2-ax^2-b(b-a)x^4+x^4\epsilon(x).$$
thus, we need
$b-a=-\frac{1}{2}$ and $b(b-a)=-\frac{1}{24}$.
which gives
$b=\frac{1}{12}$ and $a=\frac{7}{12}$.
