Prove that $$\lim_{x \rightarrow 0} \left[\frac{\arcsin x}{x}\right] = 1$$ and $$\lim_{x \rightarrow 0} \left[\frac{\arctan x}{x}\right] = 0$$ where $[\,]$ represents the greatest integer function.
For showing $\displaystyle\lim_{x \rightarrow 0} \left[\frac{\sin x}{x}\right] = 0$ and $\displaystyle\lim_{x \rightarrow 0} \left[\frac{\tan x}{x}\right] = 1$, I knew that $0 < \sin x < x < \tan x, x \in \left(0,\dfrac{\pi}{2}\right)$, hence that was easy but for $ \arcsin x$, I know no such inequality. Does $0 < \arctan x < x < \arcsin x$ hold for $x \in (0,1)$?
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$\begingroup$For the second one, consider $$f(x)=\frac {\arctan x}{x}$$
And $g(x)=\arctan x$ while $h(x)=x$
Let's find $$\lim_{x\to 0^+} \frac {\arctan x}{x}$$
On applying L'Hospital rule we get $$\lim_{x\to 0^+} \frac {\arctan x}{x}=\lim_{x\to 0^+} \frac {1}{1+x^2}\to \frac {1}{1^+}\to 1^-$$
On checking the limit of $f(x)$ as $x\to 0^-$ , we similarly get $$\lim_{x\to 0^+} \frac {\arctan x}{x}\to 1^-$$
So it is pretty easy to notice that if we apply the greatest integer function to $f(x)$ , then the value which reaches the floor function will be $1^-$ which will be converted to absolute $0$ by the floor function. Hence $$\lim_{x\to 0} [f(x)]=0$$
Try the same with the first one
$\endgroup$ 1 $\begingroup$Hint. Since $\sin(x)$ and $\tan(x)$ are increasing in $x \in (0,\pi/2)$ then $$0 < \sin x < x\quad\mbox{and}\quad x< \tan x$$ imply that for $x\in (0,1)$ $$0=\arcsin(0) < x < \arcsin(x)\quad\mbox{and}\quad \arctan(x) < x.$$ Note that you need also that $x<\sin(2x)$ in $(0,\pi/4)$ which implies that in $(0,1/\sqrt{2})$, $$\arcsin(x)<2x.$$
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