<s>[INST] Limits using derivatives If a differentiable function $f(x)$ satisfies a functional rule 
$f(x) + f(x+2) + f(x+4) = 0$ for all $x$ belonging to real numbers,
Then find the value of :
$$\lim \limits_{x \to 0} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)} $$
what's the method?
The answer is $32$.
 [/INST] If dont know if the following can help you
\begin{align}
&f(x) + f(x+2) + f(x+4) = 0\hspace{1cm} (I)\\
&f(x+2) + f(x+4) + f(x+6) = 0\hspace{1cm} (II)\\
&f(x+6) + f(x+8) + f(x+10) = 0\hspace{1cm} (III)\\
&f(x+8) + f(x+10) + f(x+12) = 0\hspace{1cm} (IV)\\
&f(x+12) + f(x+14) + f(x+16) = 0\hspace{1cm} (V)\\
&f(x+14) + f(x+16) + f(x+18) = 0\hspace{1cm} (VI)\\
\end{align}
First $(I)-(II)$ gives $f(x)=f(x+6)$ In particular $f(0)=f(6)$
also $(I)-(II)+(III)-(IV)$ gives $f(x)=f(x+12)$ In particular $f(0)=f(12)$
also also $(I)-(II)+(III)-(IV)+(V)-(VI)$ gives $f(x)=f(x+18)$ 
etc ... etc ..
Noting that $\lim \limits_{x \to 1} x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)=\frac{\pi}{4}$. So
$$\lim \limits_{x \to 1} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)}= \frac{4}{\pi} \lim \limits_{x \to 1} (f(x)+f(0))^{2}$$
</s>