Alright this looks like a very simple problem at the first go.
I need to find $$\lim_{x\rightarrow0^+} { \left\lfloor{\frac{x}{\sin (x)}}\right\rfloor}$$
So since I know the inner function's graph beforehand I know the answer will be 1.
But now here's my problem.When I'm trying to find which function $x$ or $\sin(x)$ is greater when $x \rightarrow 0^+$ I take a function $g(x)=x-\sin(x)$ and find its derivative.So $g'(x)=1-\cos(x)$.Now when I find $g'(x \rightarrow 0^+)$ I get $0$ !
So I'm not being able to prove that $g(x)$ is increasing when $x$ is slightly greater than $0$.And thus I can't prove that $x>sin(x)$ when $x$ is slighty greater than $0$ and neither can I prove $\frac{x}{\sin (x)}>1$.Where am I going wrong?