Originally Posted by

**ThePerfectHacker** I am sure you know that,

lim (x-> 0) sin x/x = 1

It can easily be proven by the rigorous definition of sine (either infinite series or solution to y''+y=0 with y(0)=0 and y'(0)=1).

However, if I would have to demonstrate this to students I would like a more simple/applied/geometric demonstration.

The one I have in my book (and another book) use the squeeze theorem. By dividing a circle sector into parts and finding the area of the circle sector. But there is a problem with this, in order to show the area of the circle is what it is we need to first show what the derivative of sin x is. And to show what the derivative of sin x is we need to know the limit of sin x/x thus it is circular (get the pun).

I have seen this version which is much more cleaner.

sin x < x <tan x

On, 0<x<pi/2

And then using the squeeze theorem appropriately.

I can show that,

sin x < x

By the following diagram below (line is shortest distance between two points).

But how do I show that,

x < tan x

In the diagram the angle is measure in radians and it is a unit circle. Thus, the intercepted arc is x and the blue line is sin x and the red line is tan x.

I can show that

blue < black (sector)

again by saying line is shortest distance.

Thus,

sin x < x

But how do I arrive at,

black (sector) < red

Thus,

x < tan x?