Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain  the 5,6,7 equation line? Thank you

While I'm not entirely sure what's going on with the answer you have provided (exponents seem to appear and disappear, multiplication turns into addition etc.), I shall offer you two methods to evaluate your given limit.

Method 1 : As both the numerator and denominator both tend to $0$ as $x\to0 $, we can use L'Hopital's Rule
$$\displaystyle{\lim_{x \to 0}}  \frac{12-6x^2-12\cos(x)}{x^4}=\displaystyle{\lim_{x \to 0}}\frac{-12x+12\sin(x)}{4x^3}$$
As both numerator and denominator still tend to $0$, we can use the rule again,
$$\displaystyle{\lim_{x \to 0}}\frac{-12x+12\sin(x)}{4x^3}=\displaystyle{\lim_{x \to 0}} \frac{-12+12\cos(x)}{12x^2}=\displaystyle{\lim_{x \to 0}}\frac{\cos(x)-1}{x^2}$$
and again,
$$\displaystyle{\lim_{x \to 0}}\frac{\cos(x)-1}{x^2}=\displaystyle{\lim_{x \to 0}}\frac{-\sin(x)}{2x}=-\frac{1}{2}$$
to get the answer.

Method 2 : We could also use the Taylor series expansion for $\cos(x)$ directly, without the cumbersome $y=2x$ substitution or the conversion of $\cos(2y)$ into $\sin^2(y)$.
$$\displaystyle{\lim_{x \to 0}}  \frac{12-6x^2-12\cos(x)}{x^4}=\displaystyle{\lim_{x \to 0}} \frac{12-6x^2-12(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}=\displaystyle{\lim_{x \to 0}}\frac{-12(\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}$$
$$\implies{\lim_{x \to 0}}  \frac{12-6x^2-12\cos(x)}{x^4}= $$$$\displaystyle{\lim_{x \to 0}}\frac{-12(\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}=-\frac{12}{4!}=-\frac{1}{2}$$

Links that may be helpful:
L'Hopital's Rule
Taylor Series of cos(x)
