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
 A: Use $\cos z=1-z^2/2!+z^4/4!+...$
$$L=\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$$$$=\displaystyle\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 (1-\frac{x^2}{2}+\frac{x^4}{24}+...)}{x^{4}}=-\frac{1}{2}.$$
A: $5th$ $step$

$1-\cos2y=\sin^2y$

$6th$ $step$

$expansion\ of\ \sin = y−\frac{y^3}{3!}+\frac{y^5}{5!}−⋯∞$

A: $$\lim_{x\to0}\frac {12-6x^2-12\cos x}{x^4}=\lim_{x\to0}\frac{-12x+12\sin x}{4x^3}=\lim_{x\to0}\frac{-12+12\cos x}{12x^2}=\lim_{x\to0}\frac{-12\sin x}{24x}=\lim_{x\to0}-1/2\cos x=-1/2$$, by repeated application of L'hopital.
A: 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)
A: You can use this transformation:
$$\lim_{x \to 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}
=24\lim _{x \rightarrow 0} \frac{\sin^2{\frac{x}{2}}-\frac{x^{2}}{4}}{x^{4}}
=\frac{24}{16}\lim _{y \rightarrow 0} \frac{\sin^2{y}-y^2}{y^{4}} 
=\frac{24}{16}\lim _{y \rightarrow 0} \left(\frac{\sin{y}-y}{y^{3}}.\frac{\sin{y}+y}{y}\right)$$
The first factor can be evaluated without using L'H rule by this link.
The second factor equals to 2.
Then your limit equals $$\frac{24}{16}.2.\frac{-1}{6}=\frac{-1}{2}$$.
Thanks for reading.
