Have fun L'hopitaling this one as well!

We make use of the following formulas, which come from Taylor expansions:

$\sin x \approx x, 1-\cos x \approx \dfrac{1}{2} x^2, \arctan x \approx x, \arcsin x \approx x, \tan x \approx x, (1+x)^a - 1 \approx ax, a \in \mathbb{R}$

$L = \displaystyle \lim_{x \to 0} \dfrac{x^2 \left[ \left( 4x + 1 + 3(1- \cos x) + \dfrac{5}{2} x^2 \right)^{2017} - 1 \right] }{2017 x (1- \cos x)}$

$L = \displaystyle \lim_{x \to 0} \dfrac{x^2 \left[ \left( 4x + 1 + \dfrac{3}{2}x^2 + \dfrac{5}{2} x^2 \right)^{2017} - 1\right]}{2017 x \dfrac{1}{2}x^2}$

$L = 2 \displaystyle \lim_{x \to 0} \dfrac{\left( 4x + 1 + 4 x^2 \right)^{2017} - 1}{2017 x }$

$L = 2 \displaystyle \lim_{x \to 0} \dfrac{ (2x+1)^{2 \cdot 2017} - 1}{2017 x }$

$L = 2 \displaystyle \lim_{x \to 0} \dfrac{2 \cdot 2 \cdot 2017 x}{2017 x } = \boxed{8}$