Help with question on limits It is given that $\lim _{ x\rightarrow 0 }{ \frac { f(x) }{ { x }^{ 2 } } =a } $ and $\lim _{ x\rightarrow 0 }{ \frac { f(1-\cos x) }{ g(x)\sin^2x } = b }$, where $b \neq 0$, then find $\lim _{ x\rightarrow 0 }{ \frac { g(1-\cos2x) }{ x^4 } }$.
My Approach:
Since $\sin^2x + \cos^2x = 1 \implies \sin^2x = (1 - \cos x)(1+\cos x)$.
$$\therefore \lim _{x \rightarrow 0} \frac{f(1-\cos x)}{g(x)(1-cos x)(1+cos x)}=b$$
$$\implies\lim _{x \rightarrow 0} \frac{f(1-\cos x)}{(1-cos x)^2} \times \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=b$$
$$\implies a\times \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=b$$
Now, since $a$ can't be equal to $0$, because if $a = 0 \implies b=0$, which is not possible as per the question, thus, 
$$\implies \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=\frac{b}{a}$$
After this, I am not able to proceed further. How should I get to the answer?
 A: You are on the right track. 
Method 1.
One may recall that, by the Taylor series expansion, as $u \to 0$, $$ \cos u=1-\frac{u^2}2+o(u^2) $$ giving, as $x \to 0$, $$ \cos 2x=1-2x^2+o(x^2) $$ and $$ \cos (1-\cos 2x)=1-2x^4+o(x^4). \tag1 $$ Now you have obtained, as $u \to 0$, $$ \frac{(1-\cos u)}{g(u)(1+\cos u)} \to \frac{b}{a} \tag2 $$ then replace $u$ with $1-\cos 2x$, as $x \to 0$ in $(2)$ to get using $(1)$, $$ \frac{(1-\cos (1-\cos 2x))}{g(1-\cos 2x)(1+\cos (1-\cos 2x))}=\frac{2x^4+o(x^4)}{g(1-\cos 2x)(2-2x^4+o(x^4))} \tag3 $$ giving, as $x \to 0$,

$$ \frac{x^4}{g(1-\cos 2x)} \sim \frac ba\tag4 $$ 

obtaining then the desired limit.
Method 2.
One may use
$$
1-\cos 2x=2\sin^2 x 
$$ giving
$$
1-\cos(1-\cos 2x)=1-\cos(2\sin^2 x )=2\sin^2(\sin^2x)
$$ then, making $x \to 0$ in $(3)$ (on the left hand side of) and writting $$
\begin{align} &\frac{(1-\cos (1-\cos 2x))}{g(1-\cos 2x)(1+\cos (1-\cos 2x))}
\\\\&=\frac{\sin^2(\sin^2x)}{(\sin^2x)^2}\cdot \frac{(\sin^2x)^2}{(x^2)^2}\cdot \frac{2}{1+\cos(2\sin^2 x )}\cdot\frac{x^4}{g(1-\cos 2x)} \tag5 
\end{align}$$ we get $(4)$ again using $\displaystyle \lim_{u\to 0}\frac{\sin u}{u}=1$.
