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.
 A: From
$$\arcsin|x|>|x|$$ and $$\arctan|x|<|x|$$ for small $|x|$, you draw
$$100\frac{\arcsin|x|}{|x|}>100$$ and
$$100\frac{\arctan|x|}{|x|}<100.$$
Then as the limits of these expressions are both $100$ you can find a neighborhood of $0$ where
$$\left\lfloor100\frac{\arcsin|x|}{|x|}\right\rfloor=100$$ and
$$\left\lfloor100\frac{\arctan|x|}{|x|}\right\rfloor=99.$$
A: Your limit is:
$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor + \lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$
Split it up:
$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor}  + \lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$
Now, lets consider the left part first:
$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor}$$
While, you're right the limit $\lim_{x\to0}{\left( \frac{\sin^{-1}{x}}{x}\right)}=1$, however the floor function can change that. 
Keep in mind that the value of $\sin^{-1}{x}$ is slightly greater than $x$, for $x>0$ and slightly less that $x$, for $x<0$. This would give us:
$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor}=100$$
Desmos® graphing calculator confirms this:

However, for
$$\lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$
The situation is different. Here, $\tan^{-1}{x}$ is less than $x$ for $x>0$, and greater than $x$, for $x<0$. This means that:
$$\lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}=99$$
Again, Desmos® gives us a good reason why:

Combining both results, we get:
$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor + \lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}=199$$
I hope this helps. Leave a comment if you didn't understand any part of the above.
A: 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 
$$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\
\lim_{x\to0}\left[100\frac{x+\frac{x^3}{6}}{x}\right]+\left[100\frac{x-\frac{x^3}{3}}{x}\right]=\\
\lim_{x\to0}\left[100(1+\frac{x^2}{6})\right]+\left[100(1-\frac{x^2}{3})\right]=\\
\lim_{x\to0}\left[100+\frac{100x^2}{6}\right]+\left[100-\frac{100x^2}{3})\right]=\\
\left[100^+\right]+\left[100^-\right]=100+99$$
A: Since $$\lim_{x\to 0}{\frac{\sin^{-1}(x)}{x}} =1$$ and $\frac{\sin^{-1}(x)}{x}\geq1$ 
$$\exists \epsilon >0 \  / \ \forall x \in ]-\epsilon,\epsilon[, 1\leq\frac{\sin^{-1}(x)}{x}<1.01 $$
then $$ \forall x \in ]-\epsilon,\epsilon[, 100\leq 100\frac{\sin^{-1}(x)}{x}<101 $$
 $$ \forall x \in ]-\epsilon,\epsilon[, \left[ 100\frac{\sin^{-1}(x)}{x}\right] = 100 $$
$$\lim_{x\to 0}{\left[ 100\frac{\sin^{-1}(x)}{x}\right]} = 100$$
Can you apply the same method to $\tan^{-1}$ ? But remember that $\frac{\tan^{-1}(x)}{x}\leq1$ 
