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
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.
sin x < x
But how do I arrive at,
black (sector) < red
x < tan x?