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?
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.
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 $\frac{2x}{\pi}<\sin x$.
First proof using mean value Theorem Let $0 < x < \pi/2$ be fixed. By Mean value theorem there exists $y \in (0, x)$ such that $$\frac{\sin(x)}{x}= \cos(y).$$ Similarly there exists $z \in (x, \pi/2)$ such that $$\frac{1-\sin(x)}{\pi/2- x}= \cos(z).$$ As $0 < y < z< \pi/2$ and $\cos$ is a strictly decreasing function in $[0, \pi/2]$ we see that $ \cos (z) < \cos (y).$ Thus $$\frac{1-\sin(x)}{\pi/2- x} < \frac{\sin(x)}{x}\implies \frac{1-\sin(x)}{\sin(x)} < \frac{\pi/2- x}{x}.$$ $$\implies\frac{1}{\sin(x)}- 1< \frac{\pi}{2x}- 1\implies \frac{2}{\pi}< \frac{\sin (x)}{x}.$$ Since $y \in (0, \pi/2)$, therefore $\frac{\sin(x)}{x}= \cos(y) <1.$ Hence for any $0 < x < \pi/2$, we get $$\frac 2 \pi<\frac{\sin(x)}{x}<1.$$
Second proof consider the function $$f(x) = \frac{\sin{x}}{x} \quad\implies f'(x) = \frac{g(x)}{x^2}~~ \text{with} \quad g(x) = x\cos{x} -\sin{x} $$ and $$g'(x) = -x\sin{x}\le 0$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, then $g$ is decreasing whereas $g(0) = 0 $.
Thus $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, hence $f$ is decreasing on $[0,\frac{\pi}{2}]$. That is $or ~~~0\le x\le \pi/2$ we have $$1=f(0)>\frac{\sin{x}}{x} =f(x) > f(\pi/2) =\frac{2}{\pi}$$
