Solving the limit: $\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$ Find the value of the limit $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$$ where $[\cdot]$ denotes the greatest integer function or the box function.
My attempt: I am aware of the standard limits $$\lim_{x\to0}\left(\frac{\sin^{-1}(x)}{x} \right) = 1 $$ and $$\lim_{x\to0}\left(\frac{\tan^{-1}(x)}{x} \right) = 1 $$ but am not sure how will I apply the box function on this limit.
Any detailed explanation to help me understand this concept will be appreciated.

When $x\to 0 $
$$\sin x\sim x-\frac{x^3}{6} \space , \space \space\space\sin^{-1} x\sim x+\frac{x^3}{6}\\\tan x\sim x+\frac{x^3}{3}\space ,\space\space \space\tan^{-1} x\sim x-\frac{x^3}{3} $$so 
$$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\
\lim_{x\to0}\left[100\frac{x+\frac{x^3}{6}}{x}\right]+\left[100\frac{x-\frac{x^3}{3}}{x}\right]=\\
\lim_{x\to0}\left[100(1+\frac{x^2}{6})\right]+\left[100(1-\frac{x^2}{3})\right]=\\
\lim_{x\to0}\left[100+\frac{100x^2}{6}\right]+\left[100-\frac{100x^2}{3})\right]=\\
\left[100^+\right]+\left[100^-\right]=100+99$$
