Limit of the floor function of $\frac{x}{\sin(x)}$

Question

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?

Answer 1

For $g(x) = x - \sin x$ we have $g'(x) = 1 - \cos x > 0$ for $0 < x \leqslant \pi/2$. and $g(0) = 0$. By the mean value theorem for any $x$ in the interval there exists $\xi$ between $0$ and $x$ such that $g(x) = g'(\xi)x > 0.$

Answer 2

There is a "classical" proof (based on geometry and trigonometry) that $\displaystyle\lim_{x\to 0} {\sin x \over x} = 1$. In it, it is proven that $\sin x < x < \tan x$ when $x>0$ and is small. It should be in any Calculus textbook.

Dividing this equation through by $\sin x$, you get $\displaystyle 1 < {x\over \sin x}$, when $x>0$ and $x$ is small. Hence $\displaystyle\left\lfloor {x\over \sin x}\right\rfloor \ge 1$ for those same $x$.