TITLE: Mean Value Theorem: $\frac{2}{\pi}<\frac{\sin x}{x}<1$
QUESTION [8 upvotes]: 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?

REPLY [9 votes]: 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.