For some holomorphic function $f$ show that $$\lim_{t\to0} \frac{cr(f(ta),f(tb),f(tc),f(td))-cr(a,b,c,d)}{t^2cr(a,b,c,d)}=\frac{(a-b)(c-d)}{6}S(f)(0)$$ where $cr(a,b,c,d)$ is the cross-ratio of four points in $\Bbb C$ and $S(f)$ is the Schwarzian derivative of $f$.
I am unsure on how to really make any progress after inserting the definition of the cross-ratio and turning it all into one fraction (rather than fractions in a fraction). I tried working from the other side but made even less progress, since I don't even know how I am supposed to get the factor $(a-b)(c-d)$ in there, since this is nowhere in the fraction.
Any help would be greatly appreciated.
Since $f$ is holomorphic, it has a series expansion $$ f(z) = z_0 + z_1 z + z_2 z^2 + z_3 z^3 + O(z^4) \tag{1}$$ at $\;z=0$. Using the definition of the Schwarzian, we get that $$ R_0 := S(f)(0) = 6(z_1z_3-z_2^2)/z_1^2. \tag{2} $$ Using cross-ratio we get that $$ cr_1:=\textrm{cr}(a,b,c,d)= \frac{(a-c)(b-d)}{(a-d)(b-c)} \tag{3} $$ and let $$ cr_2:=\textrm{cr}(f(ta),f(tb),f(tc),f(td)). \tag{4} $$ Using algebra on power series and simplifying gives us $$ \frac{cr_2-cr_1}{t^2cr_1} = \frac16 R_0(a-b)(c-d) + O(t). \tag{5} $$ Taking limits as $t\to0$ gives us the identity.
What we mean by using algebra is to use equation $(1)$ and replace $f(z)$ with a truncated power series. For exmaple, we have the following result: $$ \frac{f(ta)\!-\!f(tc)}{t(a-c)} \!=\! z_1 \!+\! (a\!+\!c)z_2t + (a^2\!+\!ac\!+\!c^2)z_3t^2 \!+\! O(t^3). \tag{6}$$ The rest of the calculations are similar, keeping only the power series terms which we can determine.