Answer
$\dfrac{2}{3}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{2 \cos 2 x-2+4x^2}{2x^4}$
Recall the series for $\cos x=\Sigma_{k =0}^{\infty} (-1)^k \dfrac{(x)^{2k}}{(2k)!}$
Now, $L=\lim\limits_{x \to 0} \dfrac{2 \Sigma_{k =0}^{\infty} (-1)^k \dfrac{(2x)^{2k}}{(2k)!} -2+4x^2}{2x^4} \\=\lim\limits_{x \to 0} [ (-1)^k \dfrac{(2)^{2k}x^{2k-4}}{(2k)!}\dfrac{1}{x^4}+\dfrac{2}{x^2}] \\=\lim\limits_{x \to 0} ( \dfrac{1}{x^4}-\dfrac{2}{x^2}+ \dfrac{2^4}{4!}-.........-\dfrac{1}{x^4}+\dfrac{2}{x^2}] \\=\lim\limits_{x \to 0} [\dfrac{2^4}{4!}-\dfrac{2^6 x^2}{6!}+.......]\\=\dfrac{2}{3}-\dfrac{2^6 (0^2)}{6!}+.......]\\=\dfrac{2}{3}$
