Answer
$\dfrac{-11}{120}$
Work Step by Step
Write the Taylor series for the function as: $\sin x=x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+......$ and $\tan^{-1} (x)=x-\dfrac{(x)^3}{3}+\dfrac{(x)^5}{5}+......$
Now, we have:
$\lim\limits_{x \to 0} \dfrac{2\sin x-\tan^{-1} x -x}{2x^5}=\lim\limits_{x \to 0} \dfrac{2(x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+.....)-(x-\dfrac{(x)^3}{3}+\dfrac{(x)^5}{5}+....) -x}{2x^5}
\\=\lim\limits_{x \to 0} [\dfrac{\dfrac{x^5}{60}-\dfrac{x^5}{5}}{2x^5}]\\=\dfrac{-11}{120}$
