_ x → 0 ^ 2 (π ^ 4 x) x ^ 4 is equal to :

Given that

$\lim_{x \to 0}\dfrac{\sin^{2}\left(\pi{\left(1 - \sin^{2}x\right)}^{2}\right)}{{x}^{4}}$
$= \lim_{x \to 0}\dfrac{\sin^{2}\left(\pi\left(1 + \sin^{4}x - 2\sin^{2}x\right)\right)}{{x}^{4}}$

$= \lim_{x \to 0}\dfrac{\sin^{2}\left(\pi - \pi\left(2\sin^{2}x - \sin^{4}x\right)\right)}{{x}^{4}}$
$= \lim_{x \to 0}\dfrac{\sin^{2}\left(\pi\left(2\sin^{2}x - \sin^{4}x\right)\right)}{{x}^{4}}$$= \lim_{x \to 0}{\left(\dfrac{\sin \left[\pi\left(2\sin^{2}x - \sin^{4}x\right)\right]}{\pi\left(2\sin^{2}x - \sin^{4}x\right)}\right)}^{2} \cdot \dfrac{{\pi}^{2}{\left(2\sin^{2}x - \sin^{4}x\right)}^{2}}{{x}^{4}}$
When $x \to 0 ⇒ 2\sin^{2}x - \sin^{4}x \to 0$

Let $k = 2\sin^{2}x - \sin^{4}x$

$= \lim_{k \to 0}{\left[\dfrac{sin\pik}{\pik}\right]}^{2} . \lim_{x \to 0}\dfrac{{\pi}^{2} . \sin^{4}x{\left(2 - \sin^{2}x\right)}^{2}}{{x}^{4}}$

We know that $\lim_{x \to 0}\dfrac{sinkx}{x} = k .$

$= \dfrac{{\pi}^{2}}{{\pi}^{2}} . {\pi}^{2} . \lim_{x \to 0}{\left[\dfrac{\sin x}{x}\right]}^{4} . \lim_{x \to 0}{\left(2 - \sin^{2}x\right)}^{2}$

$= {\pi}^{2} . 1 . {\left(2\right)}^{2}$

$= 4{\pi}^{2} .$