Mean Value Theorem: $\frac{2}{\pi}<\frac{\sin x}{x}<1$ I need to show that $\dfrac{2}{\pi}<\dfrac{\sin(x)}{x}<1$ for $0<x<\dfrac{\pi}{2}$.
I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).
We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.
Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
 A: Let $f(x) = \dfrac{\sin x}{x}$ then we know that $\lim_{x \to 0}f(x) = 1$ (its a pretty standard limit with a standard geometric proof: see this blog post under "Proof of Standard Limits"). Let us define $f(0) = 1$ to make $f(x)$ continuous on $[0, \pi/2]$. Now we can see that $$f'(x) = \dfrac{x\cos x - \sin x}{x^{2}}$$ Now we know that if $x \in (0, \pi/2)$ then we have the inequality $x < \tan x$ (see the geometric proof of this in the blog post referred earlier) so that $x\cos x < \sin x$ and thus $f'(x) < 0$ in $(0, \pi/2)$. It follows by Mean value theorem that if $x \in (0, \pi/2)$ then we have $$\frac{f(x) - f(0)}{x} = f'(c_{1}) < 0$$ so that $f(x) < f(0)$ i.e. $\dfrac{\sin x}{x} < 1$. And again $$\frac{f(\pi/2) - f(x)}{(\pi/2) - x} = f'(c_{2}) < 0$$ so that $f(\pi/2) < f(x)$ and then $\dfrac{2}{\pi} < \dfrac{\sin x}{x}$ and we are done.
A: I will prove that $\frac{2x}{\pi}<\sin x < x$,  for all $0<x<\frac{\pi}{2}$.
Proof:
First I'll use the fact that if functions $f,g$ are continuous on $[a,b)$ and diffrentiable on $(a,b)$, and satisfies $f(a)\leq g(a)$, $f'(x)\leq g'(x)$ for all $x\in (a,b)$, then $f(x)\leq g(x)$ to all $x\in [a,b)$. 
We are ready to prove $\sin x < x$ for all $x\in [0,\frac{\pi}{2}]$. Let us define $f(x)=\sin x-x$. We will get
$$f(0)=0$$
$$f'(x)=\cos x -1$$
Hence, $f(x)\leq 0$  for all $x\in [0,\frac{\pi}{2}]$. To prove that instead of "$\leq$" there is "$<$" we use Rolle's theorem. Suppose that there is $x_0\in (0,\frac{\pi}{2}]$ for which $f(x_0)=0$; then all Rolle's theorem conditions are satisfied on the closed interval $[0,\frac{\pi}{2}]$, and therefore there must be  a point $c\in (0,x_0) $ such that $f'(c)=0$. But $f'(x)=0$ iff $\cos x =1$, which then $x=\frac{\pi}{2}$, and that is a contradiction. We deduce that for all $x_0\in (0,\frac{\pi}{2}]$ there $f(x_0)<0$, and hence we have that for all $x\in (0,\frac{\pi}{2})$, we have $\sin x <x$.
Now we define $g(x)=\frac{2x}{\pi}-\sin x$. And we have that $g(0)=0$. But in this case we have some problem:
$$g'(x)=\frac{2}{\pi}-\cos x$$
When $\frac{2}{\pi}-\cos x=0$, we have $\cos x= \frac{2}{\pi}$, we deduce that there is another solution which is obtained on $(0,\frac{\pi}{2})$. 
If we look more carefully, we will see that there is only one solution for that equation, in other words the derivative is zero only in one point. Also, we notice that
$$g(\frac{\pi}{2})=\frac{2}{\pi}\frac{\pi}{2}-\sin (\frac{\pi}{2})=1-1=0$$
Again, all Rolle's theorem conditions are satisfied for $g(x)$ on the closed $[0,\frac{\pi}{2}]$ (and in every sub interval). It is impossible that for $x_0\in (0,\frac{\pi}{2})$ there $g(x_0)=0$. Now, if $x_1\in (0,\frac{\pi}{2})$ it is also impossible to have $g(x_1)>0$, because from the continuity of $g(x)$, $g(x)$ must have zero on   $(0,\frac{\pi}{2})$. Therefore, for all $x\in (0,\frac{\pi}{2})$ we have $g(x)<0$, thus $\frac{2x}{\pi}<\sin x$.
