Q:

evaluate the following limit on trigonometry

given that \begin{equation}
\lim_{y \rightarrow 0}
\frac{\sin y}{y}=1
\end{equation}
evaluate the following
\begin{equation}
\lim_{x \rightarrow 0}
\frac{2-2\cos^2 x-2 \cos x \sin ^2 x}{x^4}
\end{equation}

A:

$$\begin{align}
2-2\cos^2{x}-2\cos{x}\sin^2{x}
&=2(1-\cos^2{x})-2\cos{x}\sin^2{x} \\[6pt]
&=2 \sin^2x-2\cos x  \sin^2x \\[6pt]
&=2\sin^2{x}(1-\cos{x})\\[6pt]
&=2\sin^2{x}\cdot(2\sin^2{\dfrac{x}{2}})
\end{align}$$
so
$$\lim_{x\to 0}\dfrac{2-2\cos^2{x}-2\cos{x}\sin^2{x}}{x^4}=\lim_{x\to 0}\left(\dfrac{\sin{x}}{x}\right)^2\cdot\left(\dfrac{\sin{\dfrac{x}{2}}}{\dfrac{x}{2}}\right)^2=1$$

