# Mean Value Theorem: Real Analysis

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?

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$$\lim_{x\to 0} \frac{\sin x}{x}=\lim_{x\to0}\frac{\sin x-\sin 0}{x}=\left.\frac{\text d\sin x}{\text dx}\right|_{x=0}=\cos 0=1$$ –  Tim Ratigan Dec 7 '13 at 9:33
@Tim note that often the limit of $\sin(x)/x$ is used to evaluate $\sin '(0)$. –  pppqqq Dec 7 '13 at 10:23
@pppqqq I'm aware, but usually the derivative of sine (for some reason) is seen as more fundamental (also, this is equivalent to solving it by L'Hopital, which is common). If OP is looking for full rigor, he/she can find it here. –  Tim Ratigan Dec 7 '13 at 10:34

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.

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amazing..........:) –  user96137 Dec 8 '13 at 7:00

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 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 "$\leq$" there is "$<$" we use Role's theorem. Suppose that there is $x_0\in (0,\frac{\pi}{2}]$ for which $f(x_0)=0$; then all Role's theorem conditions satisfies on the closed interval $[0,\frac{\pi}{2}]$, and therefore there must be 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 Role's theorem conditions satisfies 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 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 $\frac{2x}{\pi}<\sin x$.

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The English is bad(!), I welcome you to edit. –  Salech Alhasov Dec 7 '13 at 12:06
"... iff $\cos x=1 \Rightarrow x=\frac{\pi}{2}$" Shouldn't x=0 for f$\prime (x)=0$? –  GTX OC Dec 7 '13 at 12:11
Yes, this was the mistake for the contradiction. I actually just asked myself that. –  user96137 Dec 8 '13 at 5:05