Answer
$\dfrac{-8}{5}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{12x-8x^3-6 \sin 2x}{x^5}$
Recall the series for $\sin x=\Sigma_{k =0}^{\infty} (-1)^k \dfrac{(x)^{2k+1}}{(2k+1)!}$
Now, $L=\lim\limits_{x \to 0} \dfrac{12x-8x^3-6 \Sigma_{k =0}^{\infty} (-1)^k \dfrac{(2x)^{2k+1}}{(2k+1)!}}{x^5}\\=\lim\limits_{x \to 0} (\dfrac{12}{x^4}-\dfrac{8}{x^2}-6[(-1)^0 2^{(2)(0)+1} \dfrac{(x)^{2(0)-4}}{(2(0)+1)!}+(-1)^1 2^{(2)(1)+1} \dfrac{(x)^{2(1)-4}}{(2(1)+1)!}+..........])\\=\lim\limits_{x \to 0} (\dfrac{12}{x^4}-\dfrac{8}{x^2}-6[\dfrac{2}{x^4}-\dfrac{2^3}{3! x^2}+.......])\\=\dfrac{-8}{5}$
