If $\underset{x\to 0}{lim} \frac{f\left(x\right)}{x^2}=a$ and  $\underset{x\to 0}{lim} \frac{f(1-\cos x)}{g\left(x\right){\sin }^{2}x}=b$  $(where b\ne 0)$, then $\underset{x\to 0}{lim} \frac{g(1-\cos 2x)}{x^4}$ is

$$\begin{gathered} \underset{x\to 0}{lim} \frac{f(1-\cos x)}{{(1-\cos x)}^2}\frac{{(1-\cos x)}^2}{g\left(x\right){\sin }^{2}x}=b \\ \Rightarrow a\underset{x\to 0}{lim}{\left(\frac{1-\cos x}{x^2}\right)}^2{\left(\frac{x}{\sin x}\right)}^2\frac{x^2}{g\left(x\right)}=b \\ \Rightarrow a\left(\frac{1}{4}\right)\underset{x\to 0}{lim}\frac{x^2}{g\left(x\right)}=b\Rightarrow \boxed{\underset{x\to 0}{lim}\frac{x^2}{g\left(x\right)}=\frac{4b}{a}} \\ \underset{x\to 0}{lim} \frac{g(1-\cos 2x)}{x^4}=\underset{x\to 0}{lim} \frac{g(1-\cos 2x)}{{(1-\cos 2x)}^2}{\left(\frac{1-\cos 2x}{x^2}\right)}^2=\frac{a}{4b}{\left(\frac{4}{2}\right)}^2=\frac{a}{b} \end{gathered}$$