# Prove |sin x|/|x| =< 1 for all x in Real Numbers.

1. Oct 25, 2011

### armoredfrog

I was looking for some help on how to start this problem. I know that we must use the Mean-Value Theorem on |sin x| to get an f '(c). But I'm having a difficult time getting an initial start past that. Any hints and tips would be most useful. I also figure that we can let f(x) = |sin x| and g(x) = |x|.

2. Oct 25, 2011

### chiro

Hello armoredfrog and welcome to the forums.

The easiest way I see is focus on the interval [-pi/2,pi/2] since anything outside this region (sin(x) is bounded by [-1,1]).

As for the interval [-pi/2,pi/2], you can use the derivative and show that the derivative of the sin(x) term is always less or equal to plus or minus 1. Since the derivative has this bound, then you can show that the function itself will also be bounded.

Also since we deal with absolute value, just split function into parts for < 0 and >= 0.

3. Oct 25, 2011

### chiro

I also forgot to mention, you have to consider the case x = 0 seperate, but there is already a result that shows this limit to be 1.

4. Oct 26, 2011

### armoredfrog

Okay, so we wouldn't need to know what the derivative of f and g is, just that they are bounded.

5. Oct 26, 2011

### chiro

If you show that the derivative is bounded, then indirectly that shows that the function values in the interval that has the domain bounded also has the function value bounded.

You could use further properties of the derivative to be more specific, but to me it seems pointless since its easy to show the absolute value of the derivative of sin(x) is less than or equal to 1, and from that we're done (since d/dx of x or -x is 1 or -1 respectively, and then take absolute values).

6. Oct 27, 2011

### lurflurf

$$\frac{\sin(x)}{x}=\frac{\sin(x)-\sin(0)}{x-0}$$

The mean value theorem is given differentiable f and g and real a and b with a<b there exist c such that
$$\frac{f'(c x)}{g'(c x)}=\frac{f(b x)-f(a x)}{g(b x)-g(a x)}$$
where a<c<b