$\begin{aligned} & \text {} \because f(\mathrm{x})=\left|\begin{array}{ccc}1 & 6+x & 36+x^2 \\ 0 & x-3 & 3 x^2-27 \\ 0 & 2 x-4 & 8 x^2-32\end{array}\right| \\ & =1\left[(x-3)\left(8 x^2-32\right)-(2 x-4)\left(3 x^2-27\right)\right]+0 \\ & =8 x^3-32 x-24 x^2+96-\left(6 x^3-54 x-12 x^2+108\right) \\ & \Rightarrow f(x)=2 x^3-12 x^2+22 x-12=2(x-1)(x-2)(x-3) \\ & \text { Now, } f(-x)=-2(x+1)(x+2)(x+3) \\ & \lim _{x \rightarrow 1} \frac{f(x)}{f(-x)}=\lim _{x \rightarrow 1} \frac{2(x-1)(x-2)(x-3)}{-2(x+1)(x+2)(x+3)}=0\end{aligned}$